Rd Sharma Solutions for Class 8 Math Chapter 6 Algebraic Expressions And Identities are provided here with simple step-by-step explanations. These solutions for Algebraic Expressions And Identities are extremely popular among Class 8 students for Math Algebraic Expressions And Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Book of Class 8 Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Solutions. All Rd Sharma Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### Question 1:

Find each of the following product:
5x2 × 4x3

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices. However, use of these laws are subject to their applicability in the given expressions.

In the present problem, to perform the multiplication, we can proceed as follows:

$5{x}^{2}×4{x}^{3}\phantom{\rule{0ex}{0ex}}=\left(5×4\right)×\left({x}^{2}×{x}^{3}\right)$
$=20{x}^{5}$                           ($\because$ ${a}^{m}×{a}^{n}={a}^{m+n}$)

Thus, the answer is $20{x}^{5}$.

#### Question 2:

Find each of the following product:
−3a2 × 4b4

To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, ${a}^{m}×{a}^{n}={a}^{m+n}$, wherever applicable.

We have:

$-3{a}^{2}×4{b}^{4}\phantom{\rule{0ex}{0ex}}=\left(-3×4\right)×\left({a}^{2}×{b}^{4}\right)\phantom{\rule{0ex}{0ex}}=-12{a}^{2}{b}^{4}$

Thus, the answer is $-12{a}^{2}{b}^{4}$.

#### Question 3:

Find each of the following product:
(−5xy) × (−3x2yz)

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, ${a}^{m}×{a}^{n}={a}^{m+n}$, wherever applicable.

We have:

Thus, the answer is $15{x}^{3}{y}^{2}z$.

#### Question 4:

Find each of the following product:
$\frac{1}{4}xy×\frac{2}{3}{x}^{2}y{z}^{2}$

To multiply algebraic expressions, we use commutative and associative laws along with the the law of indices, that is, ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\frac{1}{4}xy×\frac{2}{3}{x}^{2}y{z}^{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{4}×\frac{2}{3}\right)×\left(x×{x}^{2}\right)×\left(y×y\right)×{z}^{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{4}×\frac{2}{3}\right)×\left({x}^{1+2}\right)×\left({y}^{1+1}\right)×{z}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}{x}^{3}{y}^{2}{z}^{2}$

Thus, the answer is $\frac{1}{6}{x}^{3}{y}^{2}{z}^{2}$.

#### Question 5:

Find each of the following product:
$\left(-\frac{7}{5}x{y}^{2}z\right)×\left(\frac{13}{3}{x}^{2}y{z}^{2}\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(-\frac{7}{5}x{y}^{2}z\right)×\left(\frac{13}{3}{x}^{2}y{z}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(-\frac{7}{5}×\frac{13}{3}\right)×\left(x×{x}^{2}\right)×\left({y}^{2}×y\right)×\left(z×{z}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(-\frac{7}{5}×\frac{13}{3}\right)×\left({x}^{1+2}\right)×\left({y}^{2+1}\right)×\left({z}^{1+2}\right)\phantom{\rule{0ex}{0ex}}=-\frac{91}{15}{x}^{3}{y}^{3}{x}^{3}$

Thus, the answer is $-\frac{91}{15}{x}^{3}{y}^{3}{x}^{3}$.

#### Question 6:

Find each of the following product:
$\left(\frac{-24}{25}{x}^{3}z\right)×\left(-\frac{15}{16}x{z}^{2}y\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(-\frac{24}{25}{x}^{3}z\right)×\left(-\frac{15}{16}x{z}^{2}y\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-\frac{24}{25}\right)×\left(-\frac{15}{16}\right)\right\}×\left({x}^{3}×x\right)×\left(z×{z}^{2}\right)×y\phantom{\rule{0ex}{0ex}}=\left\{\left(-\frac{24}{25}\right)×\left(-\frac{15}{16}\right)\right\}×\left({x}^{3+1}\right)×\left({z}^{1+2}\right)×y$
$=\frac{9}{10}{x}^{4}y{z}^{3}$

Thus, the answer is $\frac{9}{10}{x}^{4}y{z}^{3}$.

#### Question 7:

Find each of the following product:
$\left(-\frac{1}{27}{a}^{2}{b}^{2}\right)×\left(\frac{9}{2}{a}^{3}{b}^{2}{c}^{2}\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., ${a}^{m}×{a}^{n}={a}^{m+n}$.
We have:

$\left(-\frac{1}{27}{a}^{2}{b}^{2}\right)×\left(\frac{9}{2}{a}^{3}{b}^{2}{c}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(-\frac{1}{27}×\frac{9}{2}\right)×\left({a}^{2}×{a}^{3}\right)×\left({b}^{2}×{b}^{2}\right)×{c}^{2}\phantom{\rule{0ex}{0ex}}=\left(-\frac{1}{27}×\frac{9}{2}\right)×\left({a}^{2+3}\right)×\left({b}^{2+2}\right)×{c}^{2}\phantom{\rule{0ex}{0ex}}=-\frac{1}{6}{a}^{5}{b}^{4}{c}^{2}$

Thus, the answer is $-\frac{1}{6}{a}^{5}{b}^{4}{c}^{2}$.

#### Question 8:

Find each of the following product:
$\left(-7xy\right)×\left(\frac{1}{4}{x}^{2}yz\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(-7xy\right)×\left(\frac{1}{4}{x}^{2}yz\right)\phantom{\rule{0ex}{0ex}}=\left(-7×\frac{1}{4}\right)×\left(x×{x}^{2}\right)×\left(y×y\right)×z\phantom{\rule{0ex}{0ex}}=\left(-7×\frac{1}{4}\right)×\left({x}^{1+2}\right)×\left({y}^{1+1}\right)×z\phantom{\rule{0ex}{0ex}}=-\frac{7}{4}{x}^{3}{y}^{2}z$

Thus, the answer is $-\frac{7}{4}{x}^{3}{y}^{2}z$.

#### Question 9:

Find each of the following product:
(7ab) × (−5ab2c× (6abc2)

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e.,  ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(7ab\right)×\left(-5a{b}^{2}c\right)×\left(6ab{c}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{7×\left(-5\right)×6\right\}×\left(a×a×a\right)×\left(b×{b}^{2}×b\right)×\left(c×{c}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{7×\left(-5\right)×6\right\}×\left({a}^{1+1+1}\right)×\left({b}^{1+2+1}\right)×\left({c}^{1+2}\right)\phantom{\rule{0ex}{0ex}}=-210{a}^{3}{b}^{4}{c}^{3}$

Thus, the answer is $-210{a}^{3}{b}^{4}{c}^{3}$.

#### Question 10:

Find each of the following product:
(−5a) × (−10a2) × (−2a3)

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., ${a}^{m}×{a}^{n}={a}^{m+n}$.
We have:

$\left(-5a\right)×\left(-10{a}^{2}\right)×\left(-2{a}^{3}\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-5\right)×\left(-10\right)×\left(-2\right)\right\}×\left(a×{a}^{2}×{a}^{3}\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-5\right)×\left(-10\right)×\left(-2\right)\right\}×\left({a}^{1+2+3}\right)\phantom{\rule{0ex}{0ex}}=-100{a}^{6}\phantom{\rule{0ex}{0ex}}$

Thus, the answer is $-100{a}^{6}$.

#### Question 11:

Find each of the following product:
(−4x2) × (−6xy2) × (−3yz2)

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(-4{x}^{2}\right)×\left(-6x{y}^{2}\right)×\left(-3y{z}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-4\right)×\left(-6\right)×\left(-3\right)\right\}×\left({x}^{2}×x\right)×\left({y}^{2}×y\right)×{z}^{2}\phantom{\rule{0ex}{0ex}}=\left\{\left(-4\right)×\left(-6\right)×\left(-3\right)\right\}×\left({x}^{2+1}\right)×\left({y}^{2+1}\right)×{z}^{2}\phantom{\rule{0ex}{0ex}}=-72{x}^{3}{y}^{3}{z}^{2}$

Thus, the answer is $-72{x}^{3}{y}^{3}{z}^{2}$.

#### Question 12:

Find each of the following product:
$\left(-\frac{2}{7}{a}^{4}\right)×\left(-\frac{3}{4}{a}^{2}b\right)×\left(-\frac{14}{5}{b}^{2}\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(-\frac{2}{7}{a}^{4}\right)×\left(-\frac{3}{4}{a}^{2}b\right)×\left(-\frac{14}{5}{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-\frac{2}{7}\right)×\left(-\frac{3}{4}\right)×\left(-\frac{14}{5}\right)\right\}×\left({a}^{4}×{a}^{2}\right)×\left(b×{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{-\left(\frac{2}{7}×\frac{3}{4}×\frac{14}{5}\right)\right\}×{a}^{4+2}×{b}^{1+2}\phantom{\rule{0ex}{0ex}}=\left\{-\left(\frac{\overline{)2}}{\overline{)7}}×\frac{3}{{\overline{)4}}_{\overline{)2}}}×\frac{{\overline{)14}}^{{\overline{)2}}^{1}}}{5}\right)\right\}×{a}^{6}×{b}^{3}\phantom{\rule{0ex}{0ex}}=-\frac{3}{5}{a}^{6}{b}^{3}$

Thus, the answer is $-\frac{3}{5}{a}^{6}{b}^{3}$.

#### Question 13:

Find each of the following product:
$\left(\frac{7}{9}a{b}^{2}\right)×\left(\frac{15}{7}a{c}^{2}b\right)×\left(-\frac{3}{5}{a}^{2}c\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(\frac{7}{9}a{b}^{2}\right)×\left(\frac{15}{7}a{c}^{2}b\right)×\left(-\frac{3}{5}{a}^{2}c\right)\phantom{\rule{0ex}{0ex}}=\left\{\frac{7}{9}×\frac{15}{7}×\left(-\frac{3}{5}\right)\right\}×\left(a×a×{a}^{2}\right)×\left({b}^{2}×b\right)×\left({c}^{2}×c\right)\phantom{\rule{0ex}{0ex}}=\left\{\frac{{\overline{)7}}^{1}}{{\overline{)9}}_{3}}×\frac{{\overline{)15}}^{3}}{\overline{)7}}×\left(-\frac{{\overline{)3}}^{1}}{\overline{)5}}\right)\right\}×\left(a×a×{a}^{2}\right)×\left({b}^{2}×b\right)×\left({c}^{2}×c\right)\phantom{\rule{0ex}{0ex}}=\left\{\frac{{\overline{)7}}^{1}}{{\overline{)9}}_{\overline{)3}}}×\frac{{\overline{)15}}^{{\overline{)3}}^{1}}}{\overline{)7}}×\left(-\frac{{\overline{)3}}^{1}}{\overline{)5}}\right)\right\}×\left({a}^{1+1+2}\right)×\left({b}^{2+1}\right)×\left({c}^{2+1}\right)\phantom{\rule{0ex}{0ex}}=-{a}^{4}{b}^{3}{c}^{3}$

Thus, the answer is $-{a}^{4}{b}^{3}{c}^{3}$.

#### Question 14:

Find each of the following product:
$\left(\frac{4}{3}{u}^{2}vw\right)×\left(-5uv{w}^{2}\right)×\left(\frac{1}{3}{v}^{2}wu\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e.,${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(\frac{4}{3}{u}^{2}vw\right)×\left(-5uv{w}^{2}\right)×\left(\frac{1}{3}{v}^{2}wu\right)\phantom{\rule{0ex}{0ex}}=\left\{\frac{4}{3}×\left(-5\right)×\frac{1}{3}\right\}×\left({u}^{2}×u×u\right)×\left(v×v×{v}^{2}\right)×\left(w×{w}^{2}×w\right)\phantom{\rule{0ex}{0ex}}=\left\{\frac{4}{3}×\left(-5\right)×\frac{1}{3}\right\}×\left({u}^{2+1+1}\right)×\left({v}^{1+1+2}\right)×\left({w}^{1+2+1}\right)\phantom{\rule{0ex}{0ex}}=-\frac{20}{9}{u}^{4}{v}^{4}{w}^{4}$

Thus, the answer is $-\frac{20}{9}{u}^{4}{v}^{4}{w}^{4}$.

#### Question 15:

Find each of the following product:
$\left(0.5x\right)×\left(\frac{1}{3}x{y}^{2}{z}^{4}\right)×\left(24{x}^{2}yz\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:
$\left(0.5x\right)×\left(\frac{1}{3}x{y}^{2}{z}^{4}\right)×\left(24{x}^{2}yz\right)\phantom{\rule{0ex}{0ex}}=\left(0.5×\frac{1}{3}×24\right)×\left(x×x×{x}^{2}\right)×\left({y}^{2}×y\right)×\left({z}^{4}×z\right)\phantom{\rule{0ex}{0ex}}=\left(0.5×\frac{1}{3}×24\right)×\left({x}^{1+1+2}\right)×\left({y}^{2+1}\right)×\left({z}^{4+1}\right)\phantom{\rule{0ex}{0ex}}=4{x}^{4}{y}^{3}{z}^{5}$

Thus, the answer is $4{x}^{4}{y}^{3}{z}^{5}$.

#### Question 16:

Find each of the following product:
$\left(\frac{4}{3}p{q}^{2}\right)×\left(-\frac{1}{4}{p}^{2}r\right)×\left(16{p}^{2}{q}^{2}{r}^{2}\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., ${a}^{m}×{a}^{n}={a}^{m+n}$.
We have:

$\left(\frac{4}{3}p{q}^{2}\right)×\left(-\frac{1}{4}{p}^{2}r\right)×\left(16{p}^{2}{q}^{2}{r}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{\frac{4}{3}×\left(-\frac{1}{4}\right)×16\right\}×\left(p×{p}^{2}×{p}^{2}\right)×\left({q}^{2}×{q}^{2}\right)×\left(r×{r}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{\frac{4}{3}×\left(-\frac{1}{4}\right)×16\right\}×\left({p}^{1+2+2}\right)×\left({q}^{2+2}\right)×\left({r}^{1+2}\right)\phantom{\rule{0ex}{0ex}}=-\frac{16}{3}{p}^{5}{q}^{4}{r}^{3}$

Thus, the answer is $-\frac{1}{3}{p}^{5}{q}^{4}{r}^{3}$.

#### Question 17:

Find each of the following product:
(2.3xy) × (0.1x) × (0.16)

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., ${a}^{m}×{a}^{n}={a}^{m+n}$.
We have:
$\left(2.3xy\right)×\left(0.1x\right)×\left(0.16\right)\phantom{\rule{0ex}{0ex}}=\left(2.3×0.1×0.16\right)×\left(x×x\right)×y\phantom{\rule{0ex}{0ex}}=\left(2.3×0.1×0.16\right)×\left({x}^{1+1}\right)×y\phantom{\rule{0ex}{0ex}}=0.0368{x}^{2}y$

Thus, the answer is $0.0368{x}^{2}y$.

#### Question 18:

Express each of the following product as a monomials and verify the result in each case for x = 1:
(3x) × (4x) × (−5x)

We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e.,​ ${a}^{m}×{a}^{n}={a}^{m+n}$.
We have:

$\left(3x\right)×\left(4x\right)×\left(-5x\right)\phantom{\rule{0ex}{0ex}}=\left\{3×4×\left(-5\right)\right\}×\left(x×x×x\right)\phantom{\rule{0ex}{0ex}}=\left\{3×4×\left(-5\right)\right\}×\left({x}^{1+1+1}\right)\phantom{\rule{0ex}{0ex}}=-60{x}^{3}$

Substituting x = 1 in LHS, we get:

Putting x = 1 in RHS, we get:

$\because$ LHS = RHS for = 1; therefore, the result is correct

Thus, the answer is $-60{x}^{3}$.

#### Question 19:

Express each of the following product as a monomials and verify the result in each case for x = 1:
(4x2) × (−3x) × $\left(\frac{4}{5}{x}^{3}\right)$

We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e.,​ ${a}^{m}×{a}^{n}={a}^{m+n}$.

We  have:

$\left(4{x}^{2}\right)×\left(-3x\right)×\left(\frac{4}{5}{x}^{3}\right)\phantom{\rule{0ex}{0ex}}=\left\{4×\left(-3\right)×\frac{4}{5}\right\}×\left({x}^{2}×x×{x}^{3}\right)\phantom{\rule{0ex}{0ex}}=\left\{4×\left(-3\right)×\frac{4}{5}\right\}×\left({x}^{2+1+3}\right)\phantom{\rule{0ex}{0ex}}=-\frac{48}{5}{x}^{6}$

$\therefore$ $\left(4{x}^{2}\right)×\left(-3x\right)×\left(\frac{4}{5}{x}^{3}\right)=-\frac{48}{5}{x}^{6}$

Substituting x = 1 in LHS, we get:

$\text{LHS}=\left(4{x}^{2}\right)×\left(-3x\right)×\left(\frac{4}{5}{x}^{3}\right)\phantom{\rule{0ex}{0ex}}=\left(4×{1}^{2}\right)×\left(-3×1\right)×\left(\frac{4}{5}×{1}^{3}\right)\phantom{\rule{0ex}{0ex}}=4×\left(-3\right)×\frac{4}{5}\phantom{\rule{0ex}{0ex}}=-\frac{48}{5}\phantom{\rule{0ex}{0ex}}$

Putting x = 1 in RHS, we get:

$\text{RHS}=-\frac{48}{5}{x}^{6}\phantom{\rule{0ex}{0ex}}=-\frac{48}{5}×{1}^{6}\phantom{\rule{0ex}{0ex}}=-\frac{48}{5}$

$\because$ LHS = RHS for = 1; therefore, the result is correct

Thus, the answer is $-\frac{48}{5}{x}^{6}$.

#### Question 20:

Express each of the following product as a monomials and verify the result in each case for x = 1:
(5x4) × (x2)3 × (2x)2

We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., .

We have:

$\therefore$

Substituting x = 1 in LHS, we get:

Put x =1 in RHS, we get:

$\because$ LHS = RHS for = 1; therefore, the result is correct.

Thus, the answer is $20{x}^{12}$.

#### Question 21:

Express each of the following product as a monomials and verify the result in each case for x = 1:
(x2)3 × (2x) × (−4x) × (5)

We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​.

We have:

${\left({x}^{2}\right)}^{3}×\left(2x\right)×\left(-4x\right)×5\phantom{\rule{0ex}{0ex}}=\left({x}^{6}\right)×\left(2x\right)×\left(-4x\right)×5\phantom{\rule{0ex}{0ex}}=\left\{2×\left(-4\right)×5\right\}×\left({x}^{6}×x×x\right)\phantom{\rule{0ex}{0ex}}=\left\{2×\left(-4\right)×5\right\}×\left({x}^{6+1+1}\right)\phantom{\rule{0ex}{0ex}}=-40{x}^{8}\phantom{\rule{0ex}{0ex}}$

$\therefore$ ${\left({x}^{2}\right)}^{3}×\left(2x\right)×\left(-4x\right)×5=-40{x}^{8}$

Substituting x = 1 in LHS, we get:​

Putting x = 1 in RHS, we get:​

$\text{RHS}=-40{x}^{8}\phantom{\rule{0ex}{0ex}}=-40{\left(1\right)}^{8}\phantom{\rule{0ex}{0ex}}=-40×1\phantom{\rule{0ex}{0ex}}=-40\phantom{\rule{0ex}{0ex}}$

$\because$ LHS = RHS for = 1; therefore, the result is correct

Thus, the answer is $-40{x}^{8}$.

#### Question 22:

Write down the product of −8x2y6 and −20xy. Verify the product for x = 2.5, y = 1.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(-8{x}^{2}{y}^{6}\right)×\left(-20xy\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-8\right)×\left(-20\right)\right\}×\left({x}^{2}×x\right)×\left({y}^{6}×y\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-8\right)×\left(-20\right)\right\}×\left({x}^{2+1}\right)×\left({y}^{6+1}\right)\phantom{\rule{0ex}{0ex}}=-160{x}^{3}{y}^{7}\phantom{\rule{0ex}{0ex}}$
$\therefore$ $\left(-8{x}^{2}{y}^{6}\right)×\left(-20xy\right)=-160{x}^{3}{y}^{7}$

Substituting x = 2.5 and y = 1 in LHS, we get:

$\text{LHS}=\left(-8{x}^{2}{y}^{6}\right)×\left(-20xy\right)\phantom{\rule{0ex}{0ex}}=\left\{-8{\left(2.5\right)}^{2}{\left(1\right)}^{6}\right\}×\left\{-20\left(2.5\right)\left(1\right)\right\}\phantom{\rule{0ex}{0ex}}=\left\{-8\left(6.25\right)\left(1\right)\right\}×\left\{-20\left(2.5\right)\left(1\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(-50\right)×\left(-50\right)\phantom{\rule{0ex}{0ex}}=2500$

Substituting x = 2.5 and y = 1 in RHS, we get:​

$\text{RHS}=-160{x}^{3}{y}^{7}\phantom{\rule{0ex}{0ex}}=-160{\left(2.5\right)}^{3}{\left(1\right)}^{7}\phantom{\rule{0ex}{0ex}}=-160\left(15.625\right)×1\phantom{\rule{0ex}{0ex}}=-2500$

Because LHS is equal to RHS, the result is correct.

Thus, the answer is $-160{x}^{3}{y}^{7}$.

#### Question 23:

Evaluate (3.2x6y3) × (2.1x2y2) when x = 1 and y = 0.5

Ans

First multiply the expressions and then substitute the values for the variables.

To multiply algebric experssions use the commutative and the associative laws along with the law of indices, ${a}^{m}×{a}^{n}={a}^{m+n}$.
We have,

$\left(3.2{x}^{6}{y}^{3}\right)×\left(2.1{x}^{2}{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(3.2×2.1\right)×\left({x}^{6}×{x}^{2}\right)×\left({y}^{3}×{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=6.72{x}^{8}{y}^{5}\phantom{\rule{0ex}{0ex}}$
Hence, $\left(3.2{x}^{6}{y}^{3}\right)×\left(2.1{x}^{2}{y}^{2}\right)=6.72{x}^{8}{y}^{5}$

Now, substitute 1 for x and 0.5  for y in the result.

$6.72{x}^{8}{y}^{5}\phantom{\rule{0ex}{0ex}}=6.72{\left(1\right)}^{8}{\left(0.5\right)}^{5}\phantom{\rule{0ex}{0ex}}=6.72×1×0.03125\phantom{\rule{0ex}{0ex}}=0.21$

Hence, the answer is $0.21$.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(3.2{x}^{6}{y}^{3}\right)×\left(2.1{x}^{2}{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(3.2×2.1\right)×\left({x}^{6}×{x}^{2}\right)×\left({y}^{3}×{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(3.2×2.1\right)×\left({x}^{6+2}\right)×\left({y}^{3+2}\right)\phantom{\rule{0ex}{0ex}}=6.72{x}^{8}{y}^{5}\phantom{\rule{0ex}{0ex}}$

$\therefore$ $\left(3.2{x}^{6}{y}^{3}\right)×\left(2.1{x}^{2}{y}^{2}\right)=6.72{x}^{8}{y}^{5}$

Substituting x = 1 and y = 0.5 in the result, we get:

$6.72{x}^{8}{y}^{5}\phantom{\rule{0ex}{0ex}}=6.72{\left(1\right)}^{8}{\left(0.5\right)}^{5}\phantom{\rule{0ex}{0ex}}=6.72×1×0.03125\phantom{\rule{0ex}{0ex}}=0.21$

Thus, the answer is $0.21$.

#### Question 24:

Find the value of (5x6) × (−1.5x2y3) × (−12xy2) when x = 1, y = 0.5.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(5{x}^{6}\right)×\left(-1.5{x}^{2}{y}^{3}\right)×\left(-12x{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{5×\left(-1.5\right)×\left(-12\right)\right\}×\left({x}^{6}×{x}^{2}×x\right)×\left({y}^{3}×{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{5×\left(-1.5\right)×\left(-12\right)\right\}×\left({x}^{6+2+1}\right)×\left({y}^{3+2}\right)\phantom{\rule{0ex}{0ex}}=90{x}^{9}{y}^{5}\phantom{\rule{0ex}{0ex}}$

$\therefore$ $\left(5{x}^{6}\right)×\left(-1.5{x}^{2}{y}^{3}\right)×\left(-12x{y}^{2}\right)=90{x}^{9}{y}^{5}$

Substituting x = 1 and y = 0.5 in the result, we get:

$90{x}^{9}{y}^{5}\phantom{\rule{0ex}{0ex}}=90{\left(1\right)}^{9}{\left(0.5\right)}^{5}\phantom{\rule{0ex}{0ex}}=90×1×0.03125\phantom{\rule{0ex}{0ex}}=2.8125$

#### Question 25:

Evaluate (2.3a5b2) × (1.2a2b2) when a = 1 and b = 0.5.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(2.3{a}^{5}{b}^{2}\right)×\left(1.2{a}^{2}{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(2.3×1.2\right)×\left({a}^{5}×{a}^{2}\right)×\left({b}^{2}×{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(2.3×1.2\right)×\left({a}^{5+2}\right)×\left({b}^{2+2}\right)\phantom{\rule{0ex}{0ex}}=2.76{a}^{7}{b}^{4}$

$\therefore$ $\left(2.3{a}^{5}{b}^{2}\right)×\left(1.2{a}^{2}{b}^{2}\right)=2.76{a}^{7}{b}^{4}$

Substituting a =1 and b = 0.5 in the result, we get:

$2.76{a}^{7}{b}^{4}\phantom{\rule{0ex}{0ex}}=2.76{\left(1\right)}^{7}{\left(0.5\right)}^{4}\phantom{\rule{0ex}{0ex}}=2.76×1×0.0625\phantom{\rule{0ex}{0ex}}=0.1725$

Thus, the answer is $0.1725$.

#### Question 26:

Evaluate (−8x2y6) × (−20xy) for x = 2.5 and y = 1.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(-8{x}^{2}{y}^{6}\right)×\left(-20xy\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-8\right)×\left(-20\right)\right\}×\left({x}^{2}×x\right)×\left({y}^{6}×y\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-8\right)×\left(-20\right)\right\}×\left({x}^{2+1}\right)×\left({y}^{6+1}\right)\phantom{\rule{0ex}{0ex}}=160{x}^{3}{y}^{7}$

$\therefore$ $\left(-8{x}^{2}{y}^{6}\right)×\left(-20xy\right)=160{x}^{3}{y}^{7}$

Substituting x = 2.5 and y = 1 in the result, we get:

$160{x}^{3}{y}^{7}\phantom{\rule{0ex}{0ex}}=160{\left(2.5\right)}^{3}{\left(1\right)}^{7}\phantom{\rule{0ex}{0ex}}=160×15.625\phantom{\rule{0ex}{0ex}}=2500$

Thus, the answer is $2500$.

#### Question 27:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
(−xy3) × (yx3) × (xy)

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(-x{y}^{3}\right)×\left(y{x}^{3}\right)×\left(xy\right)\phantom{\rule{0ex}{0ex}}=\left(-1\right)×\left(x×{x}^{3}×x\right)×\left({y}^{3}×y×y\right)\phantom{\rule{0ex}{0ex}}=\left(-1\right)×\left({x}^{1+3+1}\right)×\left({y}^{3+1+1}\right)\phantom{\rule{0ex}{0ex}}=-{x}^{5}{y}^{5}$

To verify the result, we substitute x = 1 and y = 2 in LHS; we get:

Substituting x = 1 and y = 2 in RHS, we get:​

$\text{RHS}=-{x}^{5}{y}^{5}\phantom{\rule{0ex}{0ex}}=\left(-1\right){\left(1\right)}^{5}{\left(2\right)}^{5}\phantom{\rule{0ex}{0ex}}=\left(-1\right)×1×32\phantom{\rule{0ex}{0ex}}=-32$

Because LHS is equal to RHS, the result is correct.

Thus, the answer is $-{x}^{5}{y}^{5}$.

#### Question 28:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
$\left(\frac{1}{8}{x}^{2}{y}^{4}\right)×\left(\frac{1}{4}{x}^{4}{y}^{2}\right)×\left(xy\right)×5$

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(\frac{1}{8}{x}^{2}{y}^{4}\right)×\left(\frac{1}{4}{x}^{4}{y}^{2}\right)×\left(xy\right)×5\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{8}×\frac{1}{4}×5\right)×\left({x}^{2}×{x}^{4}×x\right)×\left({y}^{4}×{y}^{2}×y\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{8}×\frac{1}{4}×5\right)×\left({x}^{2+4+1}\right)×\left({y}^{4+2+1}\right)\phantom{\rule{0ex}{0ex}}=\frac{5}{32}{x}^{7}{y}^{7}$

To verify the result, we substitute x = 1 and y = 2 in LHS; we get:

$\text{LHS}=\left(\frac{1}{8}{x}^{2}{y}^{4}\right)×\left(\frac{1}{4}{x}^{4}{y}^{2}\right)×\left(xy\right)×5\phantom{\rule{0ex}{0ex}}=\left\{\frac{1}{8}×{\left(1\right)}^{2}×{\left(2\right)}^{4}\right\}×\left\{\frac{1}{4}×{\left(1\right)}^{4}×{\left(2\right)}^{2}\right\}×\left(1×2\right)×5\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{8}×1×16\right)×\left(\frac{1}{4}×1×4\right)×\left(1×2\right)×5\phantom{\rule{0ex}{0ex}}=2×1×2×5\phantom{\rule{0ex}{0ex}}=20$

Substituting x = 1 and y = 2 in RHS, we get:​

$\text{RHS}=\frac{5}{32}{x}^{7}{y}^{7}\phantom{\rule{0ex}{0ex}}=\frac{5}{32}{\left(1\right)}^{7}{\left(2\right)}^{7}\phantom{\rule{0ex}{0ex}}=\frac{5}{\overline{)32}}×1×{\overline{)128}}^{4}\phantom{\rule{0ex}{0ex}}=20\phantom{\rule{0ex}{0ex}}$

Because LHS is equal to RHS, the result is correct.

Thus, the answer is $\frac{5}{32}{x}^{7}{y}^{7}$.

#### Question 29:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
$\left(\frac{2}{5}{a}^{2}b\right)×\left(-15{b}^{2}ac\right)×\left(-\frac{1}{2}{c}^{2}\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(\frac{2}{5}{a}^{2}b\right)×\left(-15{b}^{2}ac\right)×\left(-\frac{1}{2}{c}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{\frac{2}{5}×\left(-15\right)×\left(-\frac{1}{2}\right)\right\}×\left({a}^{2}×a\right)×\left(b×{b}^{2}\right)×\left(c×{c}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{\frac{2}{5}×\left(-15\right)×\left(-\frac{1}{2}\right)\right\}×\left({a}^{2+1}\right)×\left({b}^{1+2}\right)×\left({c}^{1+2}\right)\phantom{\rule{0ex}{0ex}}=3{a}^{3}{b}^{3}{c}^{3}$

$\because$ The expression doesn't consist of the variables x and y.

$\therefore$ The result cannot be verified for = 1 and = 2

Thus, the answer is $3{a}^{3}{b}^{3}{c}^{3}$.

#### Question 30:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
$\left(-\frac{4}{7}{a}^{2}b\right)×\left(-\frac{2}{3}{b}^{2}c\right)×\left(-\frac{7}{6}{c}^{2}a\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(-\frac{4}{7}{a}^{2}b\right)×\left(-\frac{2}{3}{b}^{2}c\right)×\left(-\frac{7}{6}{c}^{2}a\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-\frac{4}{7}\right)×\left(-\frac{2}{3}\right)×\left(-\frac{7}{6}\right)\right\}×\left({a}^{2}×a\right)×\left(b×{b}^{2}\right)×\left(c×{c}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-\frac{4}{7}\right)×\left(-\frac{2}{3}\right)×\left(-\frac{7}{6}\right)\right\}×\left({a}^{2+1}\right)×\left({b}^{1+2}\right)×\left({c}^{1+2}\right)\phantom{\rule{0ex}{0ex}}=-\frac{4}{9}{a}^{3}{b}^{3}{c}^{3}$

$\because$ The expression doesn't consist of the variables x and y.

$\therefore$ The result cannot be verified for = 1 and = 2.

Thus, the answer is $-\frac{4}{9}{a}^{3}{b}^{3}{c}^{3}$.

#### Question 31:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
$\left(\frac{4}{9}ab{c}^{3}\right)×\left(-\frac{27}{5}{a}^{3}{b}^{2}\right)×\left(-8{b}^{3}c\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(\frac{4}{9}ab{c}^{3}\right)×\left(-\frac{27}{5}{a}^{3}{b}^{2}\right)×\left(-8{b}^{3}c\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(\frac{4}{9}\right)×\left(-\frac{27}{5}\right)×\left(-8\right)\right\}×\left(a×{a}^{3}\right)×\left(b×{b}^{2}×{b}^{3}\right)×\left({c}^{3}×c\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(\frac{4}{9}\right)×\left(-\frac{27}{5}\right)×\left(-8\right)\right\}×\left({a}^{1+3}\right)×\left({b}^{1+2+3}\right)×\left({c}^{3+1}\right)\phantom{\rule{0ex}{0ex}}=\frac{96}{5}{a}^{4}{b}^{6}{c}^{4}$

Thus, the answer is $\frac{96}{5}{a}^{4}{b}^{6}{c}^{4}$.​

$\because$ The expression doesn't consist of the variables x and y.

$\therefore$ The result cannot be verified for = 1 and = 2

#### Question 32:

Evaluate each of the following when x = 2, y = −1.
$\left(2xy\right)×\left(\frac{{x}^{2}y}{4}\right)×\left({x}^{2}\right)×\left({y}^{2}\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(2xy\right)×\left(\frac{{x}^{2}y}{4}\right)×\left({x}^{2}\right)×\left({y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(2×\frac{1}{4}\right)×\left(x×{x}^{2}×{x}^{2}\right)×\left(y×y×{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(2×\frac{1}{4}\right)×\left({x}^{1+2+2}\right)×\left({y}^{1+1+2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{x}^{5}{y}^{4}$

$\therefore$ $\left(2xy\right)×\left(\frac{{x}^{2}y}{4}\right)×\left({x}^{2}\right)×\left({y}^{2}\right)=\frac{1}{2}{x}^{5}{y}^{4}$

Substituting x = 2 and y = $-$1 in the result, we get:

$\frac{1}{2}{x}^{5}{y}^{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{\left(2\right)}^{5}{\left(-1\right)}^{4}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×32×1\phantom{\rule{0ex}{0ex}}=16$

Thus, the answer is $16$.

#### Question 33:

Evaluate each of the following when x = 2, y = −1.
$\left(\frac{3}{5}{x}^{2}y\right)×\left(-\frac{15}{4}x{y}^{2}\right)×\left(\frac{7}{9}{x}^{2}{y}^{2}\right)$

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ ${a}^{m}×{a}^{n}={a}^{m+n}$.

We have:

$\left(\frac{3}{5}{x}^{2}y\right)×\left(-\frac{15}{4}x{y}^{2}\right)×\left(\frac{7}{9}{x}^{2}{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{\frac{3}{5}×\left(-\frac{15}{4}\right)×\frac{7}{9}\right\}×\left({x}^{2}×x×{x}^{2}\right)×\left(y×{y}^{2}×{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{\frac{3}{5}×\left(-\frac{15}{4}\right)×\frac{7}{9}\right\}×\left({x}^{2+1+2}\right)×\left({y}^{1+2+2}\right)\phantom{\rule{0ex}{0ex}}=-\frac{7}{4}{x}^{5}{y}^{5}$

$\therefore$ $\left(\frac{3}{5}{x}^{2}y\right)×\left(-\frac{15}{4}x{y}^{2}\right)×\left(\frac{7}{9}{x}^{2}{y}^{2}\right)=-\frac{7}{4}{x}^{5}{y}^{5}$.

Substituting x = 2 and y = $-$1 in the result, we get:

$-\frac{7}{4}{x}^{5}{y}^{5}\phantom{\rule{0ex}{0ex}}=-\frac{7}{4}{\left(2\right)}^{5}{\left(-1\right)}^{5}\phantom{\rule{0ex}{0ex}}=\left(-\frac{7}{4}\right)×32×\left(-1\right)\phantom{\rule{0ex}{0ex}}=56$

#### Question 1:

Identify the terms, their coefficients for each of the following expressions:
(i) 7x2yz − 5xy
(ii) x2 + x + 1
(iii) 3x2y2 − 5x2y2z2 + z2
(iv) 9 − ab + bcca
(v) $\frac{a}{2}+\frac{b}{2}-ab$
(vi) 0.2x 0.3xy + 0.5y

Definitions:

A term in an algebraic expression can be a constant, a variable or a product of constants and variables separated by the signs of addition (+) or subtraction ($-$) . Examples: 27, x, xyz, $\frac{1}{2}{x}^{2}yz$ etc.
The number factor of the term is called its coefficient.

(i) The expression $7{x}^{2}yz-5xy$ consists of two terms, i.e.,  $7{x}^{2}yz$ and $-5xy$.
The coefficient of $7{x}^{2}yz$ is 7 and the coefficient of $-5xy$ is $-5$.

(ii) The expression ${x}^{2}+x+1$ consists of three terms , i.e.,​ ${x}^{2}$, x and 1.
The coefficient of each term is 1.

(iii) The expression $3{x}^{2}{y}^{2}-5{x}^{2}{y}^{2}{z}^{2}+{z}^{2}$ consists of three terms , i.e.,​ $3{x}^{2}{y}^{2}$, $-5{x}^{2}{y}^{2}{z}^{2}$ and  ${z}^{2}$. The coefficient of $3{x}^{2}{y}^{2}$ is 3. The coefficient of $-5{x}^{2}{y}^{2}{z}^{2}$ is $-5$ and the coefficient of ${z}^{2}$ is 1.
(iv) The expression $9-ab+bc-ca$ consists of four terms — . The coefficient of the term 9 is 9. The coefficient of $-ab$ is $-$1. The coefficient of $bc$ is 1, and the coefficient of $-ca$ is $-$1.

(v) The expression $\frac{a}{2}+\frac{b}{2}-ab$ consists of three terms , i.e.,​ . The coefficient of $\frac{a}{2}$ is $\frac{1}{2}$. The coefficient of $\frac{b}{2}$ is $\frac{1}{2}$, and the coefficient of $-ab$ is $-$1.

(vi) The expression consists of three terms , i.e.,​ The coefficient of $0.2x$ is $0.2$. The coefficient of $-0.3xy$ is $-0.3$, and the coefficient of $0.5y$ is $0.5$.

#### Question 2:

Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any category?
(i) x + y
(ii) 1000
(iii) x + x2 + x3 + 4y4
(iv) 7 + a + 5b
(v) 2b − 3b2
(vi) 2y − 3y2 + 4y3
(vii) 5x − 4y + 3x
(viii) 4a − 15a2
(ix) xy + yz + zt + tx
(x) pqr
(xi) p2q + pq2
(xii) 2p + 2q

Definitions:

A polynomial is monomial if it has exactly one term. It is called binomial if it has exactly two non-zero terms. A polynomial ​is a trinomial if it has exactly three non-zero terms.

(i)   The polynomial $x+y$ has exactly two non zero terms , i.e.,​ x and y. Therefore, it is a binomial.
(ii)  The polynomial 1000 has exactly one term, i.e., 1000. Therefore, it is a monomial.
(iii) The polynomial $x+{x}^{2}+{x}^{3}+{x}^{4}$ has exactly four terms, i.e.,​ . Therefore, it doesn't belong to any of the categories.
(iv) The polynomial $7+a+5b$ has exactly three terms, i.e., 7, and 5b. Therefore, it is a trinomial.
(v)  The polynomial $2b-3{b}^{2}$ has exactly two terms, i.e.,​ 2b and $-3{b}^{2}$. Therefore, it is a binomial.
(vi) The polynomial $2y-3{y}^{2}+4{y}^{3}$ has exactly three terms, i.e.,​ 2y, $-3{y}^{2}$ and $4{y}^{3}$. Therefore, it is a trinomial.
(vii) The polynomial $5x-4y+3x$ has exactly three terms, i.e.,​ 5x, $-$4y and 3x. Therefore, it is a trinomial.
(viii) The polynomial $4a-15{a}^{2}$ has exactly two terms, i.e., 4a and $-15{a}^{2}$. Therefore, it is a binomial.
(ix) The polynomial $xy+yz+zt+tx$ has exactly four terms xy, yz, zt and tx. Therefore, it doesn't belong to any of the categories.
(x) The polynomial pqr has exactly one term, i.e., pqr. Therefore, it is a monomial.
(xi) The polynomial ${p}^{2}q+p{q}^{2}$ has exactly two terms, i.e.,​ ${p}^{2}q$ and $p{q}^{2}$. Therefore, it is a binomial.
(xi) The polynomial $2p+2q$ has two terms, i.e.,​ 2p and 2q. Therefore, it is a binomial.

#### Question 1:

Find the following product:
2a3(3a + 5b)

To find the product, we will use distributive law as follows:

$2{a}^{3}\left(3a+5b\right)\phantom{\rule{0ex}{0ex}}=2{a}^{3}×3a+2{a}^{3}×5b\phantom{\rule{0ex}{0ex}}=\left(2×3\right)\left({a}^{3}×a\right)+\left(2×5\right){a}^{3}b\phantom{\rule{0ex}{0ex}}=\left(2×3\right){a}^{3+1}+\left(2×5\right){a}^{3}b\phantom{\rule{0ex}{0ex}}=6{a}^{4}+10{a}^{3}b$

Thus, the answer is $6{a}^{4}+10{a}^{3}b$.

#### Question 2:

Find the following product:
−11a(3a + 2b)

To find the product, we will use distributive law as follows:

$-11a\left(3a+2b\right)\phantom{\rule{0ex}{0ex}}=\left(-11a\right)×3a+\left(-11a\right)×2b\phantom{\rule{0ex}{0ex}}=\left(-11×3\right)×\left(a×a\right)+\left(-11×2\right)×\left(a×b\right)\phantom{\rule{0ex}{0ex}}=\left(-33\right)×\left({a}^{1+1}\right)+\left(-22\right)×\left(a×b\right)\phantom{\rule{0ex}{0ex}}=-33{a}^{2}-22ab$

Thus, the answer is $-33{a}^{2}-22ab$.

#### Question 3:

Find the following product:
−5a(7a − 2b)

To find the product, we will use distributive law as follows:

$-5a\left(7a-2b\right)\phantom{\rule{0ex}{0ex}}=\left(-5a\right)×7a+\left(-5a\right)×\left(-2b\right)\phantom{\rule{0ex}{0ex}}=\left(-5×7\right)×\left(a×a\right)+\left(-5×\left(-2\right)\right)×\left(a×b\right)\phantom{\rule{0ex}{0ex}}=\left(-35\right)×\left({a}^{1+1}\right)+\left(10\right)×\left(a×b\right)\phantom{\rule{0ex}{0ex}}=-35{a}^{2}+10ab$

Thus, the answer is $-35{a}^{2}+10ab$.

#### Question 4:

Find the following product:
−11y2(3y + 7)

To find the product, we will use distributive law as follows:

$-11{y}^{2}\left(3y+7\right)\phantom{\rule{0ex}{0ex}}=\left(-11{y}^{2}\right)×3y+\left(-11{y}^{2}\right)×7\phantom{\rule{0ex}{0ex}}=\left(-11×3\right)\left({y}^{2}×y\right)+\left(-11×7\right)×\left({y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(-33\right)\left({y}^{2+1}\right)+\left(-77\right)×\left({y}^{2}\right)\phantom{\rule{0ex}{0ex}}=-33{y}^{3}-77{y}^{2}$

Thus, the answer is $-33{y}^{3}-77{y}^{2}$.

#### Question 5:

Find the following product:
$\frac{6x}{5}\left({x}^{3}+{y}^{3}\right)$

To find the product, we will use distributive law as follows:

$\frac{6x}{5}\left({x}^{3}+{y}^{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{6x}{5}×{x}^{3}+\frac{6x}{5}×{y}^{3}\phantom{\rule{0ex}{0ex}}=\frac{6}{5}×\left(x×{x}^{3}\right)+\frac{6}{5}×\left(x×{y}^{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{6}{5}×\left({x}^{1+3}\right)+\frac{6}{5}×\left(x×{y}^{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{6{x}^{4}}{5}+\frac{6x{y}^{3}}{5}$

Thus, the answer is $\frac{6{x}^{4}}{5}+\frac{6x{y}^{3}}{5}$.

#### Question 6:

xy(x3y3)

To find the product, we will use the distributive law in the following way:

$xy\left({x}^{3}-{y}^{3}\right)\phantom{\rule{0ex}{0ex}}=xy×{x}^{3}-xy×{y}^{3}\phantom{\rule{0ex}{0ex}}=\left(x×{x}^{3}\right)×y-x×\left(y×{y}^{3}\right)\phantom{\rule{0ex}{0ex}}={x}^{1+3}y-x{y}^{1+3}\phantom{\rule{0ex}{0ex}}={x}^{4}y-x{y}^{4}$

Thus, the answer is ${x}^{4}y-x{y}^{4}$.

#### Question 7:

Find the following product:
0.1y(0.1x5 + 0.1y)

To find the product, we will use distributive law as follows:

$0.1y\left(0.1{x}^{5}+0.1y\right)\phantom{\rule{0ex}{0ex}}=\left(0.1y\right)\left(0.1{x}^{5}\right)+\left(0.1y\right)\left(0.1y\right)\phantom{\rule{0ex}{0ex}}=\left(0.1×0.1\right)\left(y×{x}^{5}\right)+\left(0.1×0.1\right)\left(y×y\right)\phantom{\rule{0ex}{0ex}}=\left(0.1×0.1\right)\left({x}^{5}×y\right)+\left(0.1×0.1\right)\left({y}^{1+1}\right)\phantom{\rule{0ex}{0ex}}=0.01{x}^{5}y+0.01{y}^{2}$

Thus, the answer is $0.01{x}^{5}y+0.01{y}^{2}$.

#### Question 8:

Find the following product:
$\left(-\frac{7}{4}a{b}^{2}c-\frac{6}{25}{a}^{2}{c}^{2}\right)\left(-50{a}^{2}{b}^{2}{c}^{2}\right)$

To find the product, we will use distributive law as follows:

$\left(-\frac{7}{4}a{b}^{2}c-\frac{6}{25}{a}^{2}{c}^{2}\right)\left(-50{a}^{2}{b}^{2}{c}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-\frac{7}{4}a{b}^{2}c\right)\left(-50{a}^{2}{b}^{2}{c}^{2}\right)\right\}-\left\{\left(\frac{6}{25}{a}^{2}{c}^{2}\right)\left(-50{a}^{2}{b}^{2}{c}^{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=\left\{\left\{-\frac{7}{4}×\left(-50\right)\right\}\left(a×{a}^{2}\right)×\left({b}^{2}×{b}^{2}\right)×\left(c×{c}^{2}\right)\right\}-\left\{\left(\frac{6}{25}\right)\left(-50\right)\left({a}^{2}×{a}^{2}\right)×\left({b}^{2}\right)×\left({c}^{2}×{c}^{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=\left\{-\frac{7}{4}×\left(-50\right)\right\}\left({a}^{1+2}{b}^{2+2}{c}^{1+2}\right)-\left\{\left(\frac{6}{25}\right)\left(-50\right)\left({a}^{2+2}{b}^{2}{c}^{2+2}\right)\right\}\phantom{\rule{0ex}{0ex}}=\frac{175}{2}{a}^{3}{b}^{4}{c}^{3}-\left(-12{a}^{4}{b}^{2}{c}^{4}\right)\phantom{\rule{0ex}{0ex}}=\frac{175}{2}{a}^{3}{b}^{4}{c}^{3}+12{a}^{4}{b}^{2}{c}^{4}$

Thus, the answer is $\frac{175}{2}{a}^{3}{b}^{4}{c}^{3}+12{a}^{4}{b}^{2}{c}^{4}$.

#### Question 9:

Find the following product:
$-\frac{8}{27}xyz\left(\frac{3}{2}xy{z}^{2}-\frac{9}{4}x{y}^{2}{z}^{3}\right)$

To find the product, we will use the distributive law in the following way:

$-\frac{8}{27}xyz\left(\frac{3}{2}xy{z}^{2}-\frac{9}{4}x{y}^{2}{z}^{3}\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-\frac{8}{27}xyz\right)\left(\frac{3}{2}xy{z}^{2}\right)\right\}-\left\{\left(-\frac{8}{27}xyz\right)\left(\frac{9}{4}x{y}^{2}{z}^{3}\right)\right\}\phantom{\rule{0ex}{0ex}}=\left\{\left(-\frac{8}{27}×\frac{3}{2}\right)\left(x×x\right)×\left(y×y\right)×\left(z×{z}^{2}\right)\right\}-\left\{\left(-\frac{8}{27}×\frac{9}{4}\right)\left(x×x\right)×\left(y×{y}^{2}\right)×\left(z×{z}^{3}\right)\right\}\phantom{\rule{0ex}{0ex}}=\left\{\left(-\frac{8}{27}×\frac{3}{2}\right)\left({x}^{1+1}{y}^{1+1}{z}^{1+2}\right)\right\}-\left\{\left(-\frac{8}{27}×\frac{9}{4}\right)\left({x}^{1+1}{y}^{1+2}{z}^{1+3}\right)\right\}\phantom{\rule{0ex}{0ex}}=\left\{\left(-\frac{{\overline{)8}}^{4}}{{\overline{)27}}_{9}}×\frac{\overline{)3}}{\overline{)2}}\right)\left({x}^{1+1}{y}^{1+1}{z}^{1+2}\right)\right\}-\left\{\left(-\frac{{\overline{)8}}^{2}}{{\overline{)27}}_{3}}×\frac{\overline{)9}}{\overline{)4}}\right)\left({x}^{1+1}{y}^{1+2}{z}^{1+3}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\frac{4}{9}{x}^{2}{y}^{2}{z}^{3}+\frac{2}{3}{x}^{2}{y}^{3}{z}^{4}$

Thus, the answer is $-\frac{4}{9}{x}^{2}{y}^{2}{z}^{3}+\frac{2}{3}{x}^{2}{y}^{3}{z}^{4}$.

#### Question 10:

Find the following product:
$-\frac{4}{27}xyz\left(\frac{9}{2}{x}^{2}yz-\frac{3}{4}xy{z}^{2}\right)$

To find the product, we will use distributive law as follows:

$-\frac{4}{27}xyz\left(\frac{9}{2}{x}^{2}yz-\frac{3}{4}xy{z}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(-\frac{4}{27}xyz\right)\left(\frac{9}{2}{x}^{2}yz\right)\right\}-\left\{\left(-\frac{4}{27}xyz\right)\left(\frac{3}{4}xy{z}^{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=\left\{\left(-\frac{4}{27}×\frac{9}{2}\right)\left({x}^{1+2}{y}^{1+1}{z}^{1+1}\right)\right\}-\left\{\left(-\frac{4}{27}×\frac{3}{4}\right)\left({x}^{1+1}{y}^{1+1}{z}^{1+2}\right)\right\}\phantom{\rule{0ex}{0ex}}=\left\{\left(-\frac{{\overline{)4}}^{2}}{{\overline{)27}}_{3}}×\frac{\overline{)9}}{\overline{)2}}\right)\left({x}^{1+2}{y}^{1+1}{z}^{1+1}\right)\right\}-\left\{\left(-\frac{{\overline{)4}}^{1}}{{\overline{)27}}_{9}}×\frac{\overline{)3}}{\overline{)4}}\right)\left({x}^{1+1}{y}^{1+1}{z}^{1+2}\right)\right\}\phantom{\rule{0ex}{0ex}}=-\frac{2}{3}{x}^{3}{y}^{2}{z}^{2}+\frac{1}{9}{x}^{2}{y}^{2}{z}^{3}$

Thus, the answer is $-\frac{2}{3}{x}^{3}{y}^{2}{z}^{2}+\frac{1}{9}{x}^{2}{y}^{2}{z}^{3}$.

#### Question 11:

Find the following product:
1.5x(10x2y − 100xy2)

To find the product, we will use distributive law as follows:

$1.5x\left(10{x}^{2}y-100x{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(1.5x×10{x}^{2}y\right)-\left(1.5x×100x{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(15{x}^{1+2}y\right)-\left(150{x}^{1+1}{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=15{x}^{3}y-150{x}^{2}{y}^{2}$

Thus, the answer is $15{x}^{3}y-150{x}^{2}{y}^{2}$.

#### Question 12:

Find the following product:
4.1xy(1.1xy)

To find the product, we will use distributive law as follows:

$4.1xy\left(1.1x-y\right)\phantom{\rule{0ex}{0ex}}=\left(4.1xy×1.1x\right)-\left(4.1xy×y\right)\phantom{\rule{0ex}{0ex}}=\left\{\left(4.1×1.1\right)×xy×x\right\}-\left(4.1xy×y\right)\phantom{\rule{0ex}{0ex}}=\left(4.51{x}^{1+1}y\right)-\left(4.1x{y}^{1+1}\right)\phantom{\rule{0ex}{0ex}}=4.51{x}^{2}y-4.1x{y}^{2}$

Thus, the answer is $4.51{x}^{2}y-4.1x{y}^{2}$.

#### Question 13:

Find the following product:
250.5xy$\left(xz+\frac{y}{10}\right)$

To find the product, we will use distributive law as follows:

$250.5xy\left(xz+\frac{y}{10}\right)\phantom{\rule{0ex}{0ex}}=250.5xy×xz+250.5xy×\frac{y}{10}\phantom{\rule{0ex}{0ex}}=250.5{x}^{1+1}yz+25.05x{y}^{1+1}\phantom{\rule{0ex}{0ex}}=250.5{x}^{2}yz+25.05x{y}^{2}$

Thus, the answer is $250.5{x}^{2}yz+25.05x{y}^{2}$.

#### Question 14:

Find the following product:
$\frac{7}{5}{x}^{2}y\left(\frac{3}{5}x{y}^{2}+\frac{2}{5}x\right)$

To find the product, we will use distributive law as follows:

$\frac{7}{5}{x}^{2}y\left(\frac{3}{5}x{y}^{2}+\frac{2}{5}x\right)\phantom{\rule{0ex}{0ex}}=\frac{7}{5}{x}^{2}y×\frac{3}{5}x{y}^{2}+\frac{7}{5}{x}^{2}y×\frac{2}{5}x\phantom{\rule{0ex}{0ex}}=\frac{21}{25}{x}^{2+1}{y}^{1+2}+\frac{14}{25}{x}^{2+1}y\phantom{\rule{0ex}{0ex}}=\frac{21}{25}{x}^{3}{y}^{3}+\frac{14}{25}{x}^{3}y$

Thus, the answer is $\frac{21}{25}{x}^{3}{y}^{3}+\frac{14}{25}{x}^{3}y$.

#### Question 15:

Find the following product:

To find the product, we will use distributive law as follows:

$\frac{4}{3}a\left({a}^{2}+{b}^{2}-3{c}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{4}{3}a×{a}^{2}+\frac{4}{3}a×{b}^{2}-\frac{4}{3}a×3{c}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4}{3}{a}^{1+2}+\frac{4}{3}a{b}^{2}-4a{c}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4}{3}{a}^{3}+\frac{4}{3}a{b}^{2}-4a{c}^{2}$

Thus, the answer is $\frac{4}{3}{a}^{3}+\frac{4}{3}a{b}^{2}-4a{c}^{2}$.

#### Question 16:

Find the product 24x2 (1 − 2x) and evaluate its value for x = 3.

To find the product, we will use distributive law as follows:

$24{x}^{2}\left(1-2x\right)\phantom{\rule{0ex}{0ex}}=24{x}^{2}×1-24{x}^{2}×2x\phantom{\rule{0ex}{0ex}}=24{x}^{2}-48{x}^{1+2}\phantom{\rule{0ex}{0ex}}=24{x}^{2}-48{x}^{3}$

Substituting  x = 3 in the result, we get:

$24{x}^{2}-48{x}^{3}\phantom{\rule{0ex}{0ex}}=24{\left(3\right)}^{2}-48{\left(3\right)}^{3}\phantom{\rule{0ex}{0ex}}=24×9-48×27\phantom{\rule{0ex}{0ex}}=216-1296\phantom{\rule{0ex}{0ex}}=-1080$

Thus, the product is .

#### Question 17:

Find the product −3y(xy + y2) and find its value for x = 4 and y = 5.

To find the product, we will use distributive law as follows:​

$-3y\left(xy+{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=-3y×xy+\left(-3y\right)×{y}^{2}\phantom{\rule{0ex}{0ex}}=-3x{y}^{1+1}-3{y}^{1+2}\phantom{\rule{0ex}{0ex}}=-3x{y}^{2}-3{y}^{3}$

Substituting x = 4 and y = 5 in the result, we get:

$-3x{y}^{2}-3{y}^{3}\phantom{\rule{0ex}{0ex}}=-3\left(4\right){\left(5\right)}^{2}-3{\left(5\right)}^{3}\phantom{\rule{0ex}{0ex}}=-3\left(4\right)\left(25\right)-3\left(125\right)\phantom{\rule{0ex}{0ex}}=-300-375\phantom{\rule{0ex}{0ex}}=-675$

Thus, the product is ($-3x{y}^{2}-3{y}^{3}$), and its value for ​x = 4 and y = 5 is ($-$675).

#### Question 18:

Multiply and verify the answer for x = 1 and y = 2.

To find the product, we will use distributive law as follows:

$-\frac{3}{2}{x}^{2}{y}^{3}×\left(2x-y\right)\phantom{\rule{0ex}{0ex}}=\left(-\frac{3}{2}{x}^{2}{y}^{3}×2x\right)-\left(-\frac{3}{2}{x}^{2}{y}^{3}×y\right)\phantom{\rule{0ex}{0ex}}=\left(-3{x}^{2+1}{y}^{3}\right)-\left(-\frac{3}{2}{x}^{2}{y}^{3+1}\right)\phantom{\rule{0ex}{0ex}}=-3{x}^{3}{y}^{3}+\frac{3}{2}{x}^{2}{y}^{4}$

Substituting x = 1 and y = 2 in the result, we get:

$-3{x}^{3}{y}^{3}+\frac{3}{2}{x}^{2}{y}^{4}\phantom{\rule{0ex}{0ex}}=-3{\left(1\right)}^{3}{\left(2\right)}^{3}+\frac{3}{2}{\left(1\right)}^{2}{\left(2\right)}^{4}\phantom{\rule{0ex}{0ex}}=-3×1×8+\frac{3}{2}×1×16\phantom{\rule{0ex}{0ex}}=-24+24\phantom{\rule{0ex}{0ex}}=0$

Thus, the product is $-3{x}^{3}{y}^{3}+\frac{3}{2}{x}^{2}{y}^{4}$, and its value for ​x = 1 and y = 2 is 0.

#### Question 19:

Multiply the monomial by the binomial and find the value of each for x = −1, y = 0.25 and z = 0.05:
(i) 15y2(2 − 3x)
(ii) −3x(y2 + z2)
(iii) z2(xy)
(iv) xz(x2 + y2)

(i) To find the product, we will use distributive law as follows:

$15{y}^{2}\left(2-3x\right)\phantom{\rule{0ex}{0ex}}=15{y}^{2}×2-15{y}^{2}×3x\phantom{\rule{0ex}{0ex}}=30{y}^{2}-45x{y}^{2}$

Substituting x = $-$1 and y = 0.25 in the result, we get:

$30{y}^{2}-45x{y}^{2}\phantom{\rule{0ex}{0ex}}=30{\left(0.25\right)}^{2}-45\left(-1\right){\left(0.25\right)}^{2}\phantom{\rule{0ex}{0ex}}=30×0.0625-\left\{45×\left(-1\right)×0.0625\right\}\phantom{\rule{0ex}{0ex}}=30×0.0625-\left\{45×\left(-1\right)×0.0625\right\}\phantom{\rule{0ex}{0ex}}=1.875-\left(-2.8125\right)\phantom{\rule{0ex}{0ex}}=1.875+2.8125\phantom{\rule{0ex}{0ex}}=4.6875$

(ii) To find the product, we will use distributive law as follows:

$-3x\left({y}^{2}+{z}^{2}\right)\phantom{\rule{0ex}{0ex}}=-3x×{y}^{2}+\left(-3x\right)×{z}^{2}\phantom{\rule{0ex}{0ex}}=-3x{y}^{2}-3x{z}^{2}$

Substituting x = $-$1, y = 0.25​ and z = 0.05​ in the result, we get:

$-3x{y}^{2}-3x{z}^{2}\phantom{\rule{0ex}{0ex}}=-3\left(-1\right){\left(0.25\right)}^{2}-3\left(-1\right){\left(0.05\right)}^{2}\phantom{\rule{0ex}{0ex}}=-3\left(-1\right)\left(0.0625\right)-3\left(-1\right)\left(0.0025\right)\phantom{\rule{0ex}{0ex}}=01875+0.0075\phantom{\rule{0ex}{0ex}}=0.195$

(iii) To find the product, we will use distributive law as follows:

${z}^{2}\left(x-y\right)\phantom{\rule{0ex}{0ex}}={z}^{2}×x-{z}^{2}×y\phantom{\rule{0ex}{0ex}}=x{z}^{2}-y{z}^{2}$

Substituting x = $-$1, y = 0.25​ and z = 0.05​ in the result, we get:​

$x{z}^{2}-y{z}^{2}\phantom{\rule{0ex}{0ex}}=\left(-1\right){\left(0.05\right)}^{2}-\left(0.25\right){\left(0.05\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(-1\right)\left(0.0025\right)-\left(0.25\right)\left(0.0025\right)\phantom{\rule{0ex}{0ex}}=-0.0025-0.000625\phantom{\rule{0ex}{0ex}}=-0.003125$

(iv) To find the product, we will use distributive law as follows:

$xz\left({x}^{2}+{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=xz×{x}^{2}+xz×{y}^{2}\phantom{\rule{0ex}{0ex}}={x}^{3}z+x{y}^{2}z$

Substituting x = $-$1, y = 0.25​ and z = 0.05​ in the result, we get:​

${x}^{3}z+x{y}^{2}z\phantom{\rule{0ex}{0ex}}={\left(-1\right)}^{3}\left(0.05\right)+\left(-1\right){\left(0.25\right)}^{2}\left(0.05\right)\phantom{\rule{0ex}{0ex}}=\left(-1\right)\left(0.05\right)+\left(-1\right)\left(0.0625\right)\left(0.05\right)\phantom{\rule{0ex}{0ex}}=-0.05-0.003125\phantom{\rule{0ex}{0ex}}=-0.053125$

#### Question 20:

Simplify:
(i) 2x2(x3x) − 3x(x4 + 2x) − 2(x4 − 3x2)
(ii) x3y(x2 − 2x) + 2xy(x3x4)
(iii) 3a2 + 2(a + 2) − 3a(2a + 1)
(iv) x(x + 4) + 3x(2x2 − 1) + 4x2 + 4
(v) a(b c) − b(ca) − c(ab)
(vi) a(bc) + b(ca) + c(ab)
(vii) 4ab(ab) − 6a2(bb2) − 3b2(2a2a) + 2ab(b a)
(viii) x2(x2 + 1) − x3(x + 1) − x(x3 x)
(ix) 2a2 + 3a(1 − 2a3) + a(a + 1)
(x) a2(2a − 1) + 3a + a3 − 8
(xi) $\frac{3}{2}{x}^{2}\left({x}^{2}-1\right)+\frac{1}{4}{x}^{2}\left({x}^{2}+x\right)-\frac{3}{4}x\left({x}^{3}-1\right)$
(xii) a2b(ab2) + ab2(4ab − 2a2) − a3b(1 − 2b)
(xiii) a2b(a3a + 1) − ab(a4 − 2a2 + 2a) − b (a3a2 − 1)

(i) To simplify, we will use distributive law as follows:

$2{x}^{2}\left({x}^{3}-x\right)-3x\left({x}^{4}+2x\right)-2\left({x}^{4}-3{x}^{2}\right)\phantom{\rule{0ex}{0ex}}=2{x}^{5}-2{x}^{3}-3{x}^{5}-6{x}^{2}-2{x}^{4}+6{x}^{2}\phantom{\rule{0ex}{0ex}}=2{x}^{5}-3{x}^{5}-2{x}^{4}-2{x}^{3}-6{x}^{2}+6{x}^{2}\phantom{\rule{0ex}{0ex}}=-{x}^{5}-2{x}^{4}-2{x}^{3}$

(ii) To simplify, we will use distributive law as follows:​

${x}^{3}y\left({x}^{2}-2x\right)+2xy\left({x}^{3}-{x}^{4}\right)\phantom{\rule{0ex}{0ex}}={x}^{5}y-2{x}^{4}y+2{x}^{4}y-2{x}^{5}y\phantom{\rule{0ex}{0ex}}={x}^{5}y-2{x}^{5}y-2{x}^{4}y+2{x}^{4}y\phantom{\rule{0ex}{0ex}}=-{x}^{5}y$

(iii) To simplify, we will use distributive law as follows:​

$3{a}^{2}+2\left(a+2\right)-3a\left(2a+1\right)\phantom{\rule{0ex}{0ex}}=3{a}^{2}+2a+4-6{a}^{2}-3a\phantom{\rule{0ex}{0ex}}=3{a}^{2}-6{a}^{2}+2a-3a+4\phantom{\rule{0ex}{0ex}}=-3{a}^{2}-a+4$

(iv) To simplify, we will use distributive law as follows:

$x\left(x+4\right)+3x\left(2{x}^{2}-1\right)+4{x}^{2}+4\phantom{\rule{0ex}{0ex}}={x}^{2}+4x+6{x}^{3}-3x+4{x}^{2}+4\phantom{\rule{0ex}{0ex}}={x}^{2}+4{x}^{2}+4x-3x+6{x}^{3}+4\phantom{\rule{0ex}{0ex}}=5{x}^{2}+x+6{x}^{3}+4$

(v) To simplify, we will use distributive law as follows:​

$a\left(b-c\right)-b\left(c-a\right)-c\left(a-b\right)\phantom{\rule{0ex}{0ex}}=ab-ac-bc+ba-ca+cb\phantom{\rule{0ex}{0ex}}=ab+ba-ac-ca-bc+cb\phantom{\rule{0ex}{0ex}}=2ab-2ac$

(vi) To simplify, we will use distributive law as follows:​

(vii) To simplify, we will use distributive law as follows:​

$4ab\left(a-b\right)-6{a}^{2}\left(b-{b}^{2}\right)-3{b}^{2}\left(2{a}^{2}-a\right)+2ab\left(b-a\right)\phantom{\rule{0ex}{0ex}}=4{a}^{2}b-4a{b}^{2}-6{a}^{2}b+6{a}^{2}{b}^{2}-6{b}^{2}{a}^{2}+3{b}^{2}a+2a{b}^{2}-2{a}^{2}b\phantom{\rule{0ex}{0ex}}=4{a}^{2}b-6{a}^{2}b-2{a}^{2}b-4a{b}^{2}+3{b}^{2}a+2a{b}^{2}+6{a}^{2}{b}^{2}-6{b}^{2}{a}^{2}\phantom{\rule{0ex}{0ex}}=-4{a}^{2}b+a{b}^{2}$

(viii) To simplify, we will use distributive law as follows:​

${x}^{2}\left({x}^{2}+1\right)-{x}^{3}\left(x+1\right)-x\left({x}^{3}-x\right)\phantom{\rule{0ex}{0ex}}={x}^{4}+{x}^{2}-{x}^{4}-{x}^{3}-{x}^{4}+{x}^{2}\phantom{\rule{0ex}{0ex}}={x}^{4}-{x}^{4}-{x}^{4}-{x}^{3}+{x}^{2}+{x}^{2}\phantom{\rule{0ex}{0ex}}=-{x}^{4}-{x}^{3}+2{x}^{2}$

(ix) To simplify, we will use distributive law as follows:​

$2{a}^{2}+3a\left(1-2{a}^{3}\right)+a\left(a+1\right)\phantom{\rule{0ex}{0ex}}=2{a}^{2}+3a-6{a}^{4}+{a}^{2}+a\phantom{\rule{0ex}{0ex}}=2{a}^{2}+{a}^{2}+3a+a-6{a}^{4}\phantom{\rule{0ex}{0ex}}=3{a}^{2}+4a-6{a}^{4}$

(x) To simplify, we will use distributive law as follows:​

${a}^{2}\left(2a-1\right)+3a+{a}^{3}-8\phantom{\rule{0ex}{0ex}}=2{a}^{3}-{a}^{2}+3a+{a}^{3}-8\phantom{\rule{0ex}{0ex}}=2{a}^{3}+{a}^{3}-{a}^{2}+3a-8\phantom{\rule{0ex}{0ex}}=3{a}^{3}-{a}^{2}+3a-8$

(xi) To simplify, we will use distributive law as follows:​

$\frac{3}{2}{x}^{2}\left({x}^{2}-1\right)+\frac{1}{4}{x}^{2}\left({x}^{2}+x\right)-\frac{3}{4}x\left({x}^{3}-1\right)\phantom{\rule{0ex}{0ex}}=\frac{3}{2}{x}^{4}-\frac{3}{2}{x}^{2}+\frac{1}{4}{x}^{4}+\frac{1}{4}{x}^{3}-\frac{3}{4}{x}^{4}+\frac{3}{4}x\phantom{\rule{0ex}{0ex}}=\frac{3}{2}{x}^{4}+\frac{1}{4}{x}^{4}-\frac{3}{4}{x}^{4}+\frac{1}{4}{x}^{3}-\frac{3}{2}{x}^{2}+\frac{3}{4}x\phantom{\rule{0ex}{0ex}}=\left(\frac{6+1-3}{4}\right){x}^{4}+\frac{1}{4}{x}^{3}-\frac{3}{2}{x}^{2}+\frac{3}{4}x\phantom{\rule{0ex}{0ex}}={x}^{4}+\frac{1}{4}{x}^{3}-\frac{3}{2}{x}^{2}+\frac{3}{4}x$

(xii) To simplify, we will use distributive law as follows:

${a}^{2}b\left(a-{b}^{2}\right)+a{b}^{2}\left(4ab-2{a}^{2}\right)-{a}^{3}b\left(1-2b\right)\phantom{\rule{0ex}{0ex}}={a}^{3}b-{a}^{2}{b}^{3}+4{a}^{2}{b}^{3}-2{a}^{3}{b}^{2}-{a}^{3}b+2{a}^{3}{b}^{2}\phantom{\rule{0ex}{0ex}}={a}^{3}b-{a}^{3}b-{a}^{2}{b}^{3}+4{a}^{2}{b}^{3}-2{a}^{3}{b}^{2}+2{a}^{3}{b}^{2}\phantom{\rule{0ex}{0ex}}=3{a}^{2}{b}^{3}$

(xiii) To simplify, we will use distributive law as follows:​

${a}^{2}b\left({a}^{3}-a+1\right)-ab\left({a}^{4}-2{a}^{2}+2a\right)-b\left({a}^{3}-{a}^{2}-1\right)\phantom{\rule{0ex}{0ex}}={a}^{5}b-{a}^{3}b+{a}^{2}b-{a}^{5}b+2{a}^{3}b-2{a}^{2}b-{a}^{3}b+{a}^{2}b+b\phantom{\rule{0ex}{0ex}}={a}^{5}b-{a}^{5}b-{a}^{3}b+2{a}^{3}b-{a}^{3}b+{a}^{2}b-2{a}^{2}b+{a}^{2}b+b\phantom{\rule{0ex}{0ex}}=b$

#### Question 1:

Multiply:
(5x + 3) by (7x + 2)

To multiply, we will use distributive law as follows:

$\left(5x+3\right)\left(7x+2\right)\phantom{\rule{0ex}{0ex}}=5x\left(7x+2\right)+3\left(7x+2\right)\phantom{\rule{0ex}{0ex}}=\left(5x×7x+5x×2\right)+\left(3×7x+3×2\right)\phantom{\rule{0ex}{0ex}}=\left(35{x}^{2}+10x\right)+\left(21x+6\right)\phantom{\rule{0ex}{0ex}}=35{x}^{2}+10x+21x+6\phantom{\rule{0ex}{0ex}}=35{x}^{2}+31x+6$

Thus, the answer is $35{x}^{2}+31x+6$.

#### Question 2:

Multiply:
(2x + 8) by (x − 3)

To multiply the expressions, we will use the distributive law in the following way:

$\left(2x+8\right)\left(x-3\right)\phantom{\rule{0ex}{0ex}}=2x\left(x-3\right)+8\left(x-3\right)\phantom{\rule{0ex}{0ex}}=\left(2x×x-2x×3\right)+\left(8x-8×3\right)\phantom{\rule{0ex}{0ex}}=\left(2{x}^{2}-6x\right)+\left(8x-24\right)\phantom{\rule{0ex}{0ex}}=2{x}^{2}-6x+8x-24\phantom{\rule{0ex}{0ex}}=2{x}^{2}+2x-24$

Thus, the answer is $2{x}^{2}+2x-24$.

#### Question 3:

Multiply:
(7x + y) by (x + 5y)

To multiply, we will use distributive law as follows:

$\left(7x+y\right)\left(x+5y\right)\phantom{\rule{0ex}{0ex}}=7x\left(x+5y\right)+y\left(x+5y\right)\phantom{\rule{0ex}{0ex}}=7{x}^{2}+35xy+xy+5{y}^{2}\phantom{\rule{0ex}{0ex}}=7{x}^{2}+36xy+5{y}^{2}$

Thus, the answer is $7{x}^{2}+36xy+5{y}^{2}$.

#### Question 4:

Multiply:
(a − 1) by (0.1a2 + 3)

To multiply, we will use distributive law as follows:

$\left(a-1\right)\left(0.1{a}^{2}+3\right)\phantom{\rule{0ex}{0ex}}=0.1{a}^{2}\left(a-1\right)+3\left(a-1\right)\phantom{\rule{0ex}{0ex}}=0.1{a}^{3}-0.1{a}^{2}+3a-3$

Thus, the answer is $0.1{a}^{3}-0.1{a}^{2}+3a-3$.

#### Question 5:

Multiply:
(3x2 + y2) by (2x2 + 3y2)

To multiply, we will use distributive law as follows:

$\left(3{x}^{2}+{y}^{2}\right)\left(2{x}^{2}+3{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=3{x}^{2}\left(2{x}^{2}+3{y}^{2}\right)+{y}^{2}\left(2{x}^{2}+3{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=6{x}^{4}+9{x}^{2}{y}^{2}+2{x}^{2}{y}^{2}+3{y}^{4}\phantom{\rule{0ex}{0ex}}=6{x}^{4}+11{x}^{2}{y}^{2}+3{y}^{4}$

Thus, the answer is $6{x}^{4}+11{x}^{2}{y}^{2}+3{y}^{4}$.

#### Question 6:

Multiply:

To multiply, we will use distributive law as follows:

$\left(\frac{3}{5}x+\frac{1}{2}y\right)\left(\frac{5}{6}x+4y\right)\phantom{\rule{0ex}{0ex}}=\frac{3}{5}x\left(\frac{5}{6}x+4y\right)+\frac{1}{2}y\left(\frac{5}{6}x+4y\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{x}^{2}+\frac{12}{5}xy+\frac{5}{12}xy+2{y}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{x}^{2}+\left(\frac{144+25}{60}\right)xy+2{y}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{x}^{2}+\frac{169}{60}xy+2{y}^{2}$

Thus, the answer is $\frac{1}{2}{x}^{2}+\frac{169}{60}xy+2{y}^{2}$.

#### Question 7:

Multiply:
(x6y6) by (x2 + y2)

To multiply, we will use distributive law as follows:

$\left({x}^{6}-{y}^{6}\right)\left({x}^{2}+{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{6}\left({x}^{2}+{y}^{2}\right)-{y}^{6}\left({x}^{2}+{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left({x}^{8}+{x}^{6}{y}^{2}\right)-\left({y}^{6}{x}^{2}+{y}^{8}\right)\phantom{\rule{0ex}{0ex}}={x}^{8}+{x}^{6}{y}^{2}-{y}^{6}{x}^{2}-{y}^{8}$

Thus, the answer is ${x}^{8}+{x}^{6}{y}^{2}-{y}^{6}{x}^{2}-{y}^{8}$.

#### Question 8:

Multiply:
(x2 + y2) by (3a + 2b)

To multiply, we will use distributive law as follows:

$\left({x}^{2}+{y}^{2}\right)\left(3a+2b\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(3a+2b\right)+{y}^{2}\left(3a+2b\right)\phantom{\rule{0ex}{0ex}}=3a{x}^{2}+2b{x}^{2}+3a{y}^{2}+2b{y}^{2}$

Thus, the answer is $3a{x}^{2}+2b{x}^{2}+3a{y}^{2}+2b{y}^{2}$.

#### Question 9:

Multiply:
[−3d + (−7f)] by (5d + f)

To multiply, we will use distributive law as follows:

$\left[-3d+\left(-7f\right)\right]\left(5d+f\right)\phantom{\rule{0ex}{0ex}}=\left(-3d\right)\left(5d+f\right)+\left(-7f\right)\left(5d+f\right)\phantom{\rule{0ex}{0ex}}=\left(-15{d}^{2}-3df\right)+\left(-35df-7{f}^{2}\right)\phantom{\rule{0ex}{0ex}}=-15{d}^{2}-3df-35df-7{f}^{2}\phantom{\rule{0ex}{0ex}}=-15{d}^{2}-38df-7{f}^{2}\phantom{\rule{0ex}{0ex}}$

Thus, the answer is $-15{d}^{2}-38df-7{f}^{2}$.

#### Question 10:

Multiply:
(0.8a − 0.5b) by (1.5a − 3b)

To multiply, we will use distributive law as follows:

$\left(0.8a-0.5b\right)\left(1.5a-3b\right)\phantom{\rule{0ex}{0ex}}=0.8a\left(1.5a-3b\right)-0.5b\left(1.5a-3b\right)\phantom{\rule{0ex}{0ex}}=1.2{a}^{2}-2.4ab-0.75ab+1.5{b}^{2}\phantom{\rule{0ex}{0ex}}=1.2{a}^{2}-3.15ab+1.5{b}^{2}\phantom{\rule{0ex}{0ex}}$

Thus, the answer is $1.2{a}^{2}-3.15ab+1.5{b}^{2}$.

#### Question 11:

Multiply:
(2x2y2 − 5xy2) by (x2y2)

To multiply, we will use distributive law as follows:

$\left(2{x}^{2}{y}^{2}-5x{y}^{2}\right)\left({x}^{2}-{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=2{x}^{2}{y}^{2}\left({x}^{2}-{y}^{2}\right)-5x{y}^{2}\left({x}^{2}-{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=2{x}^{4}{y}^{2}-2{x}^{2}{y}^{4}-5{x}^{3}{y}^{2}+5x{y}^{4}\phantom{\rule{0ex}{0ex}}$

Thus, the answer is $2{x}^{4}{y}^{2}-2{x}^{2}{y}^{4}-5{x}^{3}{y}^{2}+5x{y}^{4}$.

#### Question 12:

Multiply:
$\left(\frac{x}{7}+\frac{{x}^{2}}{2}\right)by\left(\frac{2}{5}+\frac{9x}{4}\right)$

To multiply the expressions, we will use the distributive law in the following way:

Thus, the answer is  $\frac{2x}{35}+\frac{73{x}^{2}}{140}+\frac{9{x}^{3}}{8}$

#### Question 13:

Multiply:
$\left(-\frac{a}{7}+\frac{{a}^{2}}{9}\right)by\left(\frac{b}{2}-\frac{{b}^{2}}{3}\right)$

To multiply, we will use distributive law as follows:

$\left(-\frac{a}{7}+\frac{{a}^{2}}{9}\right)\left(\frac{b}{2}-\frac{{b}^{2}}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(-\frac{a}{7}\right)\left(\frac{b}{2}-\frac{{b}^{2}}{3}\right)+\left(\frac{{a}^{2}}{9}\right)\left(\frac{b}{2}-\frac{{b}^{2}}{3}\right)\phantom{\rule{0ex}{0ex}}=\left(-\frac{ab}{14}+\frac{a{b}^{2}}{21}\right)+\left(\frac{{a}^{2}b}{18}-\frac{{a}^{2}{b}^{2}}{27}\right)\phantom{\rule{0ex}{0ex}}=-\frac{ab}{14}+\frac{a{b}^{2}}{21}+\frac{{a}^{2}b}{18}-\frac{{a}^{2}{b}^{2}}{27}\phantom{\rule{0ex}{0ex}}$

Thus, the answer is $-\frac{ab}{14}+\frac{a{b}^{2}}{21}+\frac{{a}^{2}b}{18}-\frac{{a}^{2}{b}^{2}}{27}$.

#### Question 14:

Multiply:
(3x2y − 5xy2) by

To multiply, we will use distributive law as follows:

$\left(3{x}^{2}y-5x{y}^{2}\right)\left(\frac{1}{5}{x}^{2}+\frac{1}{3}{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{5}{x}^{2}\left(3{x}^{2}y-5x{y}^{2}\right)+\frac{1}{3}{y}^{2}\left(3{x}^{2}y-5x{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{3}{5}{x}^{4}y-{x}^{3}{y}^{2}+{x}^{2}{y}^{3}-\frac{5}{3}x{y}^{4}$

Thus, the answer is $\frac{3}{5}{x}^{4}y-{x}^{3}{y}^{2}+{x}^{2}{y}^{3}-\frac{5}{3}x{y}^{4}$.

#### Question 15:

Multiply:
(2x2 − 1) by (4x3 + 5x2)

To multiply, we will use distributive law as follows:

$\left(2{x}^{2}-1\right)\left(4{x}^{3}+5{x}^{2}\right)\phantom{\rule{0ex}{0ex}}=2{x}^{2}\left(4{x}^{3}+5{x}^{2}\right)-1\left(4{x}^{3}+5{x}^{2}\right)\phantom{\rule{0ex}{0ex}}=8{x}^{5}+10{x}^{4}-4{x}^{3}-5{x}^{2}$

Thus, the answer is $8{x}^{5}+10{x}^{4}-4{x}^{3}-5{x}^{2}$.

#### Question 16:

(2xy + 3y2) (3y2 − 2)

To multiply, we will use distributive law as follows:

$\left(2xy+3{y}^{2}\right)\left(3{y}^{2}-2\right)\phantom{\rule{0ex}{0ex}}=2xy\left(3{y}^{2}-2\right)+3{y}^{2}\left(3{y}^{2}-2\right)\phantom{\rule{0ex}{0ex}}=6x{y}^{3}-4xy+9{y}^{4}-6{y}^{2}\phantom{\rule{0ex}{0ex}}=9{y}^{4}+6x{y}^{3}-6{y}^{2}-4xy$

Thus, the answer is $9{y}^{4}+6x{y}^{3}-6{y}^{2}-4xy$.

#### Question 17:

Find the following product and verify the result for x = − 1, y = − 2:
(3x − 5y) (x + y)

To multiply, we will use distributive law as follows:

$\left(3x-5y\right)\left(x+y\right)\phantom{\rule{0ex}{0ex}}=3x\left(x+y\right)-5y\left(x+y\right)\phantom{\rule{0ex}{0ex}}=3{x}^{2}+3xy-5xy-5{y}^{2}\phantom{\rule{0ex}{0ex}}=3{x}^{2}-2xy-5{y}^{2}$

$\therefore$ $\left(3x-5y\right)\left(x+y\right)=3{x}^{2}-2xy-5{y}^{2}$.
Now, we put x = $-$1 and y = $-$2 on both sides to verify the result.

$\text{LHS}=\left(3x-5y\right)\left(x+y\right)\phantom{\rule{0ex}{0ex}}=\left\{3\left(-1\right)-5\left(-2\right)\right\}\left\{-1+\left(-2\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(-3+10\right)\left(-3\right)\phantom{\rule{0ex}{0ex}}=\left(7\right)\left(-3\right)\phantom{\rule{0ex}{0ex}}=-21$

$\text{RHS}=3{x}^{2}-2xy-5{y}^{2}\phantom{\rule{0ex}{0ex}}=3{\left(-1\right)}^{2}-2\left(-1\right)\left(-2\right)-5{\left(-2\right)}^{2}\phantom{\rule{0ex}{0ex}}=3×1-4-5×4\phantom{\rule{0ex}{0ex}}=3-4-20\phantom{\rule{0ex}{0ex}}=-21\phantom{\rule{0ex}{0ex}}$

Because LHS is equal to RHS, the result is verified.

Thus, the answer is $3{x}^{2}-2xy-5{y}^{2}$.

#### Question 18:

Find the following product and verify the result for x = − 1, y = − 2:
(x2y  1) (3  2x2y)

To multiply, we will  use distributive law as follows:

$\left({x}^{2}y-1\right)\left(3-2{x}^{2}y\right)\phantom{\rule{0ex}{0ex}}={x}^{2}y\left(3-2{x}^{2}y\right)-1×\left(3-2{x}^{2}y\right)\phantom{\rule{0ex}{0ex}}=3{x}^{2}y-2{x}^{4}{y}^{2}-3+2{x}^{2}y\phantom{\rule{0ex}{0ex}}=5{x}^{2}y-2{x}^{4}{y}^{2}-3$

$\therefore$ $\left({x}^{2}y-1\right)\left(3-2{x}^{2}y\right)=5{x}^{2}y-2{x}^{4}{y}^{2}-3$

Now, we put x = $-$1 and y = $-$2 on both sides to verify the result.

$\text{RHS}=5{x}^{2}y-2{x}^{4}{y}^{2}-3\phantom{\rule{0ex}{0ex}}=5{\left(-1\right)}^{2}\left(-2\right)-2{\left(-1\right)}^{4}{\left(-2\right)}^{2}-3\phantom{\rule{0ex}{0ex}}=\left[5×1×\left(-2\right)\right]-\left[2×1×4\right]-3\phantom{\rule{0ex}{0ex}}=-10-8-3\phantom{\rule{0ex}{0ex}}=-21$

Because LHS is equal to RHS, the result is verified.

Thus, the answer is $5{x}^{2}y-2{x}^{4}{y}^{2}-3$.

#### Question 19:

Find the following product and verify the result for x = − 1, y = − 2:
$\left(\frac{1}{3}x-\frac{{y}^{2}}{5}\right)\left(\frac{1}{3}x+\frac{{y}^{2}}{5}\right)$

To multiply, we will use distributive law as follows:

$\left(\frac{1}{3}x-\frac{{y}^{2}}{5}\right)\left(\frac{1}{3}x+\frac{{y}^{2}}{5}\right)\phantom{\rule{0ex}{0ex}}=\left[\frac{1}{3}x\left(\frac{1}{3}x+\frac{{y}^{2}}{5}\right)\right]-\left[\frac{{y}^{2}}{5}\left(\frac{1}{3}x+\frac{{y}^{2}}{5}\right)\right]\phantom{\rule{0ex}{0ex}}=\left[\frac{1}{9}{x}^{2}+\frac{x{y}^{2}}{15}\right]-\left[\frac{x{y}^{2}}{15}+\frac{{y}^{4}}{25}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{9}{x}^{2}+\frac{x{y}^{2}}{15}-\frac{x{y}^{2}}{15}-\frac{{y}^{4}}{25}\phantom{\rule{0ex}{0ex}}=\frac{1}{9}{x}^{2}-\frac{{y}^{4}}{25}$

$\therefore$ $\left(\frac{1}{3}x-\frac{{y}^{2}}{5}\right)\left(\frac{1}{3}x+\frac{{y}^{2}}{5}\right)=\frac{1}{9}{x}^{2}-\frac{{y}^{4}}{25}$

Now, we will put x = $-$1 and y = $-$2 on both the sides to verify the result.

$\text{RHS}=\frac{1}{9}{x}^{2}-\frac{{y}^{4}}{25}\phantom{\rule{0ex}{0ex}}=\frac{1}{9}{\left(-1\right)}^{2}-\frac{{\left(-2\right)}^{4}}{25}\phantom{\rule{0ex}{0ex}}=\frac{1}{9}×1-\frac{16}{25}\phantom{\rule{0ex}{0ex}}=\frac{1}{9}-\frac{16}{25}\phantom{\rule{0ex}{0ex}}=-\frac{119}{225}$

Because LHS is equal to RHS, the result is verified.

Thus, the answer is $\frac{1}{9}{x}^{2}-\frac{{y}^{4}}{25}$.

#### Question 20:

Simplify:
x2(x + 2y) (x − 3y)

To simplify, we will proceed as follows:

${x}^{2}\left(x+2y\right)\left(x-3y\right)\phantom{\rule{0ex}{0ex}}=\left[{x}^{2}\left(x+2y\right)\right]\left(x-3y\right)\phantom{\rule{0ex}{0ex}}=\left({x}^{3}+2{x}^{2}y\right)\left(x-3y\right)\phantom{\rule{0ex}{0ex}}={x}^{3}\left(x-3y\right)+2{x}^{2}y\left(x-3y\right)\phantom{\rule{0ex}{0ex}}={x}^{4}-3{x}^{3}y+2{x}^{3}y-6{x}^{2}{y}^{2}\phantom{\rule{0ex}{0ex}}={x}^{4}-{x}^{3}y-6{x}^{2}{y}^{2}$

Thus, the answer is ${x}^{4}-{x}^{3}y-6{x}^{2}{y}^{2}$.

#### Question 21:

Simplify:
(x2 − 2y2) (x + 4y) x2y2

To simplify, we will proceed as follows:

$\left({x}^{2}-2{y}^{2}\right)\left(x+4y\right){x}^{2}{y}^{2}\phantom{\rule{0ex}{0ex}}=\left[{x}^{2}\left(x+4y\right)-2{y}^{2}\left(x+4y\right)\right]{x}^{2}{y}^{2}\phantom{\rule{0ex}{0ex}}=\left({x}^{3}+4{x}^{2}y-2x{y}^{2}-8{y}^{3}\right){x}^{2}{y}^{2}\phantom{\rule{0ex}{0ex}}={x}^{5}{y}^{2}+4{x}^{4}{y}^{3}-2{x}^{3}{y}^{4}-8{x}^{2}{y}^{5}$

Thus, the answer is ${x}^{5}{y}^{2}+4{x}^{4}{y}^{3}-2{x}^{3}{y}^{4}-8{x}^{2}{y}^{5}$.

#### Question 22:

Simplify:
a2b2(a + 2b)(3a + b)

To simplify, we will proceed as follows:

${a}^{2}{b}^{2}\left(a+2b\right)\left(3a+b\right)\phantom{\rule{0ex}{0ex}}=\left[{a}^{2}{b}^{2}\left(a+2b\right)\right]\left(3a+b\right)\phantom{\rule{0ex}{0ex}}=\left({a}^{3}{b}^{2}+2{a}^{2}{b}^{3}\right)\left(3a+b\right)\phantom{\rule{0ex}{0ex}}=3a\left({a}^{3}{b}^{2}+2{a}^{2}{b}^{3}\right)+b\left({a}^{3}{b}^{2}+2{a}^{2}{b}^{3}\right)\phantom{\rule{0ex}{0ex}}=3{a}^{4}{b}^{2}+6{a}^{3}{b}^{3}+{a}^{3}{b}^{3}+2{a}^{2}{b}^{4}\phantom{\rule{0ex}{0ex}}=3{a}^{4}{b}^{2}+7{a}^{3}{b}^{3}+2{a}^{2}{b}^{4}$

Thus, the answer is $3{a}^{4}{b}^{2}+7{a}^{3}{b}^{3}+2{a}^{2}{b}^{4}$.

#### Question 23:

Simplify:
x2(x − y) y2(x + 2y)

To simplify, we will proceed as follows:

${x}^{2}\left(x-y\right){y}^{2}\left(x+2y\right)\phantom{\rule{0ex}{0ex}}=\left[{x}^{2}\left(x-y\right)\right]\left[{y}^{2}\left(x+2y\right)\right]\phantom{\rule{0ex}{0ex}}=\left({x}^{3}-{x}^{2}y\right)\left(x{y}^{2}+2{y}^{3}\right)\phantom{\rule{0ex}{0ex}}={x}^{3}\left(x{y}^{2}+2{y}^{3}\right)-{x}^{2}y\left(x{y}^{2}+2{y}^{3}\right)\phantom{\rule{0ex}{0ex}}={x}^{4}{y}^{2}+2{x}^{3}{y}^{3}-\left[{x}^{3}{y}^{3}+2{x}^{2}{y}^{4}\right]\phantom{\rule{0ex}{0ex}}={x}^{4}{y}^{2}+2{x}^{3}{y}^{3}-{x}^{3}{y}^{3}-2{x}^{2}{y}^{4}\phantom{\rule{0ex}{0ex}}={x}^{4}{y}^{2}+{x}^{3}{y}^{3}-2{x}^{2}{y}^{4}$

Thus, the answer is ${x}^{4}{y}^{2}+{x}^{3}{y}^{3}-2{x}^{2}{y}^{4}$.

#### Question 24:

Simplify:
(x3 − 2x2 + 5x − 7)(2x − 3)

To simplify, we will proceed as follows:

$\left({x}^{3}-2{x}^{2}+5x-7\right)\left(2x-3\right)\phantom{\rule{0ex}{0ex}}=2x\left({x}^{3}-2{x}^{2}+5x-7\right)-3\left({x}^{3}-2{x}^{2}+5x-7\right)\phantom{\rule{0ex}{0ex}}=2{x}^{4}-4{x}^{3}+10{x}^{2}-14x-3{x}^{3}+6{x}^{2}-15x+21$
$=2{x}^{4}-4{x}^{3}-3{x}^{3}+10{x}^{2}+6{x}^{2}-14x-15x+21$     (Rearranging)
$=2{x}^{4}-7{x}^{3}+16{x}^{2}-29x+21$                              (Combining like terms)

Thus, the answer is $2{x}^{4}-7{x}^{3}+16{x}^{2}-29x+21$.

#### Question 25:

Simplify:
(5x + 3)(x − 1)(3x − 2)

To simplify, we will proceed as follows:

$\left(5x+3\right)\left(x-1\right)\left(3x-2\right)\phantom{\rule{0ex}{0ex}}=\left[\left(5x+3\right)\left(x-1\right)\right]\left(3x-2\right)$
$=\left[5x\left(x-1\right)+3\left(x-1\right)\right]\left(3x-2\right)$              (Distributive law)
$=\left[5{x}^{2}-5x+3x-3\right]\left(3x-2\right)\phantom{\rule{0ex}{0ex}}=\left[5{x}^{2}-2x-3\right]\left(3x-2\right)\phantom{\rule{0ex}{0ex}}=3x\left(5{x}^{2}-2x-3\right)-2\left(5{x}^{2}-2x-3\right)\phantom{\rule{0ex}{0ex}}=15{x}^{3}-6{x}^{2}-9x-\left[10{x}^{2}-4x-6\right]\phantom{\rule{0ex}{0ex}}=15{x}^{3}-6{x}^{2}-9x-10{x}^{2}+4x+6$
$=15{x}^{3}-6{x}^{2}-10{x}^{2}-9x+4x+6$              (Rearranging)
$=15{x}^{3}-16{x}^{2}-5x+6$                              (Combining like terms)

Thus, the answer is $15{x}^{3}-16{x}^{2}-5x+6$.

#### Question 26:

Simplify:
(5 − x)(6 − 5x)( 2 − x)

To simplify, we will proceed as follows:

$\left(5-x\right)\left(6-5x\right)\left(2-x\right)\phantom{\rule{0ex}{0ex}}=\left[\left(5-x\right)\left(6-5x\right)\right]\left(2-x\right)$
$=\left[5\left(6-5x\right)-x\left(6-5x\right)\right]\left(2-x\right)$                (Distributive law)
$=\left(30-25x-6x+5{x}^{2}\right)\left(2-x\right)\phantom{\rule{0ex}{0ex}}=\left(30-31x+5{x}^{2}\right)\left(2-x\right)\phantom{\rule{0ex}{0ex}}=2\left(30-31x+5{x}^{2}\right)-x\left(30-31x+5{x}^{2}\right)\phantom{\rule{0ex}{0ex}}=60-62x+10{x}^{2}-30x+31{x}^{2}-5{x}^{3}$
$=60-62x-30x+10{x}^{2}+31{x}^{2}-5{x}^{3}$              (Rearranging)
$=60-92x+41{x}^{2}-5{x}^{3}$                                (Combining like terms)

Thus, the answer is $60-92x+41{x}^{2}-5{x}^{3}$.

#### Question 27:

Simplify:
(2x2 + 3x − 5)(3x2 − 5x + 4)

To simplify, we will proceed as follows:

$\left(2{x}^{2}+3x-5\right)\left(3{x}^{2}-5x+4\right)$
$=2{x}^{2}\left(3{x}^{2}-5x+4\right)+3x\left(3{x}^{2}-5x+4\right)-5\left(3{x}^{2}-5x+4\right)$           (Distributive law)
$=6{x}^{4}-10{x}^{3}+8{x}^{2}+9{x}^{3}-15{x}^{2}+12x-15{x}^{2}+25x-20$
$=6{x}^{4}-10{x}^{3}+9{x}^{3}+8{x}^{2}-15{x}^{2}-15{x}^{2}+12x+25x-20$              (Rearranging)
$=6{x}^{4}-{x}^{3}-22{x}^{2}+36x-20$                                                     (Combining like terms)

Thus, the answer is $6{x}^{4}-{x}^{3}-22{x}^{2}+36x-20$.

#### Question 28:

Simplify:
(3x − 2)(2x − 3) + (5x − 3)(x + 1)

To simplify, we will proceed as follows:

$\left(3x-2\right)\left(2x-3\right)+\left(5x-3\right)\left(x+1\right)\phantom{\rule{0ex}{0ex}}=\left[\left(3x-2\right)\left(2x-3\right)\right]+\left[\left(5x-3\right)\left(x+1\right)\right]$
$=\left[3x\left(2x-3\right)-2\left(2x-3\right)\right]+\left[5x\left(x+1\right)-3\left(x+1\right)\right]$           (Distributive law)
$=6{x}^{2}-9x-4x+6+5{x}^{2}+5x-3x-3$
$=6{x}^{2}+5{x}^{2}-9x-4x+5x-3x-3+6$                                  (Rearranging)
$=11{x}^{2}-11x+3$                                                                 (Combining like terms)

Thus, the answer is $11{x}^{2}-11x+3$.

#### Question 29:

Simplify:
(5x − 3)(x + 2) − (2x + 5)(4x − 3)

To simplify, we will proceed as follows:

$\left(5x-3\right)\left(x+2\right)-\left(2x+5\right)\left(4x-3\right)\phantom{\rule{0ex}{0ex}}=\left[\left(5x-3\right)\left(x+2\right)\right]-\left[\left(2x+5\right)\left(4x-3\right)\right]$
$=\left[5x\left(x+2\right)-3\left(x+2\right)\right]-\left[2x\left(4x-3\right)+5\left(4x-3\right)\right]$            (Distributive law)
$=5{x}^{2}+10x-3x-6-8{x}^{2}+6x-20x+15$
$=5{x}^{2}-8{x}^{2}+10x-3x+6x-20x-6+15$                              (Rearranging)
$=5{x}^{2}-8{x}^{2}+10x-3x+6x-20x-6+15\phantom{\rule{0ex}{0ex}}=-3{x}^{2}-7x+9$                              (Combining like terms)

Hence, the answer is $-3{x}^{2}-7x+9$.

#### Question 30:

Simplify:
(3x + 2y)(4x + 3y) − (2xy)(7x − 3y)

To simplify, we will proceed as follows:

$\left(3x+2y\right)\left(4x+3y\right)-\left(2x-y\right)\left(7x-3y\right)\phantom{\rule{0ex}{0ex}}=\left[\left(3x+2y\right)\left(4x+3y\right)\right]-\left[\left(2x-y\right)\left(7x-3y\right)\right]$
$=\left[3x\left(4x+3y\right)+2y\left(4x+3y\right)\right]-\left[2x\left(7x-3y\right)-y\left(7x-3y\right)\right]$            (Distributive law)
$=12{x}^{2}+9xy+8xy+6{y}^{2}-\left[14{x}^{2}-6xy-7xy+3{y}^{2}\right]\phantom{\rule{0ex}{0ex}}=12{x}^{2}+9xy+8xy+6{y}^{2}-14{x}^{2}+6xy+7xy-3{y}^{2}$
$=12{x}^{2}-14{x}^{2}+9xy+8xy+6xy+7xy+6{y}^{2}-3{y}^{2}$                              (Rearranging)
$=-2{x}^{2}+30xy+3{y}^{2}$                                                                       (Combining like terms)

Thus, the answer is $-2{x}^{2}+30xy+3{y}^{2}$.

#### Question 31:

Simplify:
(x2 − 3x + 2)(5x − 2) − (3x2 + 4x − 5)(2x − 1)

To simplify, we will  proceed as follows:

$\left({x}^{2}-3x+2\right)\left(5x-2\right)-\left(3{x}^{2}+4x-5\right)\left(2x-1\right)\phantom{\rule{0ex}{0ex}}=\left[\left({x}^{2}-3x+2\right)\left(5x-2\right)\right]-\left[\left(3{x}^{2}+4x-5\right)\left(2x-1\right)\right]$
$=\left[5x\left({x}^{2}-3x+2\right)-2\left({x}^{2}-3x+2\right)\right]-\left[2x\left(3{x}^{2}+4x-5\right)-1×\left(3{x}^{2}+4x-5\right)\right]$            (Distributive law)
$=\left[5{x}^{3}-15{x}^{2}+10x-\left(2{x}^{2}-6x+4\right)\right]-\left[6{x}^{3}+8{x}^{2}-10x-3{x}^{2}-4x+5\right]\phantom{\rule{0ex}{0ex}}=\left[5{x}^{3}-15{x}^{2}+10x-2{x}^{2}+6x-4\right]-\left[6{x}^{3}+8{x}^{2}-10x-3{x}^{2}-4x+5\right]\phantom{\rule{0ex}{0ex}}=5{x}^{3}-15{x}^{2}+10x-2{x}^{2}+6x-4-6{x}^{3}-8{x}^{2}+10x+3{x}^{2}+4x-5$
$=5{x}^{3}-6{x}^{3}-15{x}^{2}-2{x}^{2}-8{x}^{2}+3{x}^{2}+10x+6x+10x+4x-5-4$                                (Rearranging)
$=-{x}^{3}-22{x}^{2}+30x-9$                                                                                            (Combining like terms)

Thus, the answer is $-{x}^{3}-22{x}^{2}+30x-9$.

#### Question 32:

Simplify:
(x3 − 2x2 + 3x − 4) (x −1) − (2x − 3)(x2x + 1)

To simplify,we will proceed as follows:

$\left({x}^{3}-2{x}^{2}+3x-4\right)\left(x-1\right)-\left(2x-3\right)\left({x}^{2}-x+1\right)\phantom{\rule{0ex}{0ex}}=\left[\left({x}^{3}-2{x}^{2}+3x-4\right)\left(x-1\right)\right]-\left[\left(2x-3\right)\left({x}^{2}-x+1\right)\right]$
$=\left[x\left({x}^{3}-2{x}^{2}+3x-4\right)-1\left({x}^{3}-2{x}^{2}+3x-4\right)\right]-\left[2x\left({x}^{2}-x+1\right)-3\left({x}^{2}-x+1\right)\right]$            (Distributive law)
$=\left[x\left({x}^{3}-2{x}^{2}+3x-4\right)-1\left({x}^{3}-2{x}^{2}+3x-4\right)\right]-\left[2x\left({x}^{2}-x+1\right)-3\left({x}^{2}-x+1\right)\right]\phantom{\rule{0ex}{0ex}}={x}^{4}-2{x}^{3}+3{x}^{2}-4x-{x}^{3}+2{x}^{2}-3x+4-\left[2{x}^{3}-2{x}^{2}+2x-3{x}^{2}+3x-3\right]\phantom{\rule{0ex}{0ex}}={x}^{4}-2{x}^{3}+3{x}^{2}-4x-{x}^{3}+2{x}^{2}-3x+4-2{x}^{3}+2{x}^{2}-2x+3{x}^{2}-3x+3$
$={x}^{4}-2{x}^{3}-2{x}^{3}-{x}^{3}+3{x}^{2}+2{x}^{2}+2{x}^{2}+3{x}^{2}-4x-3x-2x-3x+4+3$                             (Rearranging)
$={x}^{4}-5{x}^{3}+10{x}^{2}-12x+7$                                                                                            (Combining like terms)

Thus, the answer is ${x}^{4}-5{x}^{3}+10{x}^{2}-12x+7$.

#### Question 1:

Write the following squares of binomials as trinomials:
(i) (x + 2)2
(ii) (8a + 3b)2
(iii) (2m + 1)2
(iv) ${\left(9a+\frac{1}{6}\right)}^{2}$
(v) ${\left(x+\frac{{x}^{2}}{2}\right)}^{2}$
(vi) $\left(\frac{x}{4}-\frac{y}{3}\right)$
(vii) ${\left(3x-\frac{1}{3x}\right)}^{2}$
(viii) ${\left(\frac{x}{y}-\frac{y}{x}\right)}^{2}$
(ix) ${\left(\frac{3a}{2}-\frac{5b}{4}\right)}^{2}$
(x) (a2bbc2)2
(xi) ${\left(\frac{2a}{3b}+\frac{2b}{3a}\right)}^{2}$
(xii) (x2ay)2

We will use the identities to convert the squares of binomials as trinomials.

#### Question 2:

Find the product of the following binomials:
(i) (2x + y)(2x + y)
(ii) (a + 2b)(a − 2b)
(iii) (a2 + bc)(a2 bc)
(iv) $\left(\frac{4x}{5}-\frac{3y}{4}\right)\left(\frac{4x}{5}+\frac{3y}{4}\right)$
(v) $\left(2x+\frac{3}{y}\right)\left(2x-\frac{3}{y}\right)$
(vi) (2a3 + b3)(2a3b3)
(vii) $\left({x}^{4}+\frac{2}{{x}^{2}}\right)\left({x}^{4}-\frac{2}{{x}^{2}}\right)$
(viii) $\left({x}^{3}+\frac{1}{{x}^{3}}\right)\left({x}^{3}-\frac{1}{{x}^{3}}\right)$

(i) We will use the identity ${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$  in the given expression to find the product.
$\left(2x+y\right)\left(2x+y\right)\phantom{\rule{0ex}{0ex}}={\left(2x+y\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(2x\right)}^{2}+2\left(2x\right)\left(y\right)+{y}^{2}\phantom{\rule{0ex}{0ex}}=4{x}^{2}+4xy+{y}^{2}$

(ii) We will use the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$ in the given expression to find the product.
$\left(a+2b\right)\left(a-2b\right)\phantom{\rule{0ex}{0ex}}={a}^{2}-{\left(2b\right)}^{2}\phantom{\rule{0ex}{0ex}}={a}^{2}-4{b}^{2}$

(iii) We will use the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$ in the given expression to find the product.
$\left({a}^{2}+bc\right)\left({a}^{2}-bc\right)\phantom{\rule{0ex}{0ex}}={\left({a}^{2}\right)}^{2}-{\left(bc\right)}^{2}\phantom{\rule{0ex}{0ex}}={a}^{4}-{b}^{2}{c}^{2}$

(iv)We will use the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$ in the given expression to find the product.
$\left(\frac{4x}{5}-\frac{3y}{4}\right)\left(\frac{4x}{5}+\frac{3y}{4}\right)\phantom{\rule{0ex}{0ex}}={\left(\frac{4x}{5}\right)}^{2}-{\left(\frac{3y}{4}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{16{x}^{2}}{25}-\frac{9{y}^{2}}{16}$

(v) We will use the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$ in the given expression to find the product.
$\left(2x+\frac{3}{y}\right)\left(2x-\frac{3}{y}\right)\phantom{\rule{0ex}{0ex}}={\left(2x\right)}^{2}-{\left(\frac{3}{y}\right)}^{2}\phantom{\rule{0ex}{0ex}}=4{x}^{2}-\frac{9}{{y}^{2}}$

(vi) We will use the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$ in the given expression to find the product.
$\left(2{a}^{3}+{b}^{3}\right)\left(2{a}^{3}-{b}^{3}\right)\phantom{\rule{0ex}{0ex}}={\left(2{a}^{3}\right)}^{2}-{\left({b}^{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=4{a}^{6}-{b}^{6}$

(vii) We will use the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$ in the given expression to find the product.
$\left({x}^{4}+\frac{2}{{x}^{2}}\right)\left({x}^{4}-\frac{2}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}={\left({x}^{4}\right)}^{2}-{\left(\frac{2}{{x}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}={x}^{8}-\frac{4}{{x}^{4}}$

(viii) We will use the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$ in the given expression to find the product.

$\left({x}^{3}+\frac{1}{{x}^{3}}\right)\left({x}^{3}-\frac{1}{{x}^{3}}\right)\phantom{\rule{0ex}{0ex}}={\left({x}^{3}\right)}^{2}-{\left(\frac{1}{{x}^{3}}\right)}^{2}\phantom{\rule{0ex}{0ex}}={x}^{6}-\frac{1}{{x}^{6}}$

#### Question 3:

Using the formula for squaring a binomial, evaluate the following:
(i) (102)2
(ii) (99)2
(iii) (1001)2
(iv) (999)2
(v) (703)2

(i) Here, we will use the identity ${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$
${\left(102\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(100+2\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(100\right)}^{2}+2×100×2+{2}^{2}\phantom{\rule{0ex}{0ex}}=10000+400+4\phantom{\rule{0ex}{0ex}}=10404$

(ii) Here, we will use the identity ${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$
${\left(99\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(100-1\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(100\right)}^{2}-2×100×1+{1}^{2}\phantom{\rule{0ex}{0ex}}=10000-200+1\phantom{\rule{0ex}{0ex}}=9801$

(iii) Here, we will use the identity ${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$
${\left(1001\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(1000+1\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(1000\right)}^{2}+2×1000×1+{1}^{2}\phantom{\rule{0ex}{0ex}}=1000000+2000+1\phantom{\rule{0ex}{0ex}}=1002001$

(iv) Here, we will use the identity ${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$
${\left(999\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(1000-1\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(1000\right)}^{2}-2×1000×1+{1}^{2}\phantom{\rule{0ex}{0ex}}=1000000-2000+1\phantom{\rule{0ex}{0ex}}=998001$

(v) Here, we will use the identity ${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$
${\left(703\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(700+3\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(700\right)}^{2}+2×700×3+{3}^{2}\phantom{\rule{0ex}{0ex}}=490000+4200+9\phantom{\rule{0ex}{0ex}}=494209$

#### Question 4:

Simplify the following using the formula: (ab)(a + b) = a2b2:
(i) (82)2 − (18)2
(ii) (467)2 − (33)2
(iii) (79)2 − (69)2
(iv) 197 × 203
(v) 113 × 87
(vi) 95 × 105
(vii) 1.8 × 2.2
(viii) 9.8 × 10.2

Here, we will use the identity

(i) Let us consider the following expression:

${\left(82\right)}^{2}-{\left(18\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(82+18\right)\left(82-18\right)\phantom{\rule{0ex}{0ex}}=100×64\phantom{\rule{0ex}{0ex}}=6400$

(ii) Let us consider the following expression:

${\left(467\right)}^{2}-{\left(33\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(467+33\right)\left(467-33\right)\phantom{\rule{0ex}{0ex}}=500×434\phantom{\rule{0ex}{0ex}}=217000$

(iii) Let us consider the following expression:

${\left(79\right)}^{2}-{\left(69\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(79+69\right)\left(79-69\right)\phantom{\rule{0ex}{0ex}}=148×10\phantom{\rule{0ex}{0ex}}=1480$

(iv) Let us consider the following product:

$197×203$

; therefore, we will write the above product as:
$197×203\phantom{\rule{0ex}{0ex}}=\left(200-3\right)\left(200+3\right)\phantom{\rule{0ex}{0ex}}={\left(200\right)}^{2}-{\left(3\right)}^{2}\phantom{\rule{0ex}{0ex}}=40000-9\phantom{\rule{0ex}{0ex}}=39991$

Thus, the answer is $39991$.

(v) Let us consider the following product:

$113×87$

$\because \frac{113+87}{2}=\frac{200}{2}=100$; therefore, we will write the above product as:
$113×87\phantom{\rule{0ex}{0ex}}=\left(100+13\right)\left(100-13\right)\phantom{\rule{0ex}{0ex}}={\left(100\right)}^{2}-{\left(13\right)}^{2}\phantom{\rule{0ex}{0ex}}=10000-169\phantom{\rule{0ex}{0ex}}=9831$

(vi) Let us consider the following product:

$95×105$

$\because \frac{95+105}{2}=\frac{200}{2}=100$; therefore, we will write the above product as:
$95×105\phantom{\rule{0ex}{0ex}}=\left(100+5\right)\left(100-5\right)\phantom{\rule{0ex}{0ex}}={\left(100\right)}^{2}-{\left(5\right)}^{2}\phantom{\rule{0ex}{0ex}}=10000-25\phantom{\rule{0ex}{0ex}}=9975$

(vii) Let us consider the following product:

$1.8×2.2$

$\because \frac{1.8+2.2}{2}=\frac{4}{2}=2$; therefore, we will write the above product as:
$1.8×2.2\phantom{\rule{0ex}{0ex}}=\left(2-0.2\right)\left(2+0.2\right)\phantom{\rule{0ex}{0ex}}={\left(2\right)}^{2}-{\left(0.2\right)}^{2}\phantom{\rule{0ex}{0ex}}=4-0.04\phantom{\rule{0ex}{0ex}}=3.96$

(viii) Let us consider the following product:

$9.8×10.2$

$\because \frac{9.8+10.2}{2}=\frac{20}{2}=10$; therefore, we will write the above product as:
$9.8×10.2\phantom{\rule{0ex}{0ex}}=\left(10-0.2\right)\left(10+0.2\right)\phantom{\rule{0ex}{0ex}}={\left(10\right)}^{2}-{\left(0.2\right)}^{2}\phantom{\rule{0ex}{0ex}}=100-0.04\phantom{\rule{0ex}{0ex}}=99.96$

#### Question 5:

Simplify the following using the identities:
(i) $\frac{{58}^{2}-{42}^{2}}{16}$
(ii) 178 × 178 − 22 × 22
(iii) $\frac{198×198-102×102}{96}$
(iv) 1.73 × 1.73 − 0.27 × 0.27
(v) $\frac{8.63×8.63-1.37×1.37}{0.726}$

(i) Let us consider the following expression:

$\frac{{58}^{2}-{42}^{2}}{16}$
Using the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$, we get:

$\frac{{58}^{2}-{42}^{2}}{16}=\frac{\left(58+42\right)\left(58-42\right)}{16}$
$⇒\frac{{58}^{2}-{42}^{2}}{16}=\frac{100×16}{16}\phantom{\rule{0ex}{0ex}}⇒\frac{{58}^{2}-{42}^{2}}{16}=100$

(ii) Let us consider the following expression:

$178×178-22×22$
Using the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$, we get:

$178×178-22×22={178}^{2}-{22}^{2}=\left(178+22\right)\left(178-22\right)=200×156=31200$

(iii) Let us consider the following expression:

$\frac{198×198-102×102}{96}=\frac{{198}^{2}-{102}^{2}}{96}$
Using the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$, we get:

$\frac{198×198-102×102}{96}=\frac{{198}^{2}-{102}^{2}}{96}=\frac{\left(198+102\right)\left(198-102\right)}{96}$
$⇒\frac{198×198-102×102}{96}=\frac{\left(198+102\right)\left(198-102\right)}{96}\phantom{\rule{0ex}{0ex}}⇒\frac{198×198-102×102}{96}=\frac{300×96}{96}\phantom{\rule{0ex}{0ex}}⇒\frac{198×198-102×102}{96}=300$

(iv) Let us consider the following expression:

$1.73×1.73-0.27×0.27$
Using the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$, we get:

$1.73×1.73-0.27×0.27=1.{73}^{2}-0.{27}^{2}=\left(1.73+0.27\right)\left(1.73-0.27\right)=2×1.46=2.92$

(v) Let us consider the following expression:

$\frac{8.63×8.63-1.37×1.37}{0.726}=\frac{8.{63}^{2}-1.{37}^{2}}{0.726}$
Using the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$, we get:

$\frac{8.63×8.63-1.37×1.37}{0.726}=\frac{8.{63}^{2}-1.{37}^{2}}{0.726}=\frac{\left(8.63+1.37\right)\left(8.63-1.37\right)}{0.726}$
$⇒\frac{8.63×8.63-1.37×1.37}{0.726}=\frac{\left(8.63+1.37\right)\left(8.63-1.37\right)}{0.726}\phantom{\rule{0ex}{0ex}}⇒\frac{8.63×8.63-1.37×1.37}{0.726}=\frac{\left(8.63+1.37\right)\left(8.63-1.37\right)}{0.726}\phantom{\rule{0ex}{0ex}}⇒\frac{8.63×8.63-1.37×1.37}{0.726}=\frac{10×7.26}{0.726}\phantom{\rule{0ex}{0ex}}⇒\frac{8.63×8.63-1.37×1.37}{0.726}=\frac{10×{\overline{)7.26}}^{10}}{\overline{)0.726}}\phantom{\rule{0ex}{0ex}}⇒\frac{8.63×8.63-1.37×1.37}{0.726}=100$

#### Question 6:

Find the value of x, if:
(i) 4x = (52)2 − (48)2
(ii) 14x = (47)2 − (33)2
(iii) 5x = (50)2 − (40)2

(i) Let us consider the following equation:

$4x={\left(52\right)}^{2}-{\left(48\right)}^{2}$
Using the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$, we get:
$4x={\left(52\right)}^{2}-{\left(48\right)}^{2}\phantom{\rule{0ex}{0ex}}4x=\left(52+48\right)\left(52-48\right)\phantom{\rule{0ex}{0ex}}4x=100×4=400$
$⇒4x=400$
$⇒x=100$        (Dividing both sides by 4)

(ii) Let us consider the following equation:

$14x={\left(47\right)}^{2}-{\left(33\right)}^{2}$
Using the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$, we get:
$14x={\left(47\right)}^{2}-{\left(33\right)}^{2}\phantom{\rule{0ex}{0ex}}14x=\left(47+33\right)\left(47-33\right)\phantom{\rule{0ex}{0ex}}14x=80×14=1120$
$⇒14x=1120$
$⇒x=80$        (Dividing both sides by 14)

(iii) Let us consider the following equation:

$5x={\left(50\right)}^{2}-{\left(40\right)}^{2}$
Using the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$, we get:
$5x={\left(50\right)}^{2}-{\left(40\right)}^{2}\phantom{\rule{0ex}{0ex}}5x=\left(50+40\right)\left(50-40\right)\phantom{\rule{0ex}{0ex}}5x=90×10=900$
$⇒5x=900$
$⇒x=180$        (Dividing both sides by 5)

#### Question 7:

If $x+\frac{1}{x}=20,$ find the value of ${x}^{2}+\frac{1}{{x}^{2}}.$

Let us consider the following equation:

$x+\frac{1}{x}=20$
Squaring both sides, we get:

$⇒{x}^{2}+\frac{1}{{x}^{2}}=398$                               (Subtracting 2 from both sides)

#### Question 8:

If $x-\frac{1}{x}=3,$ find the values of ${x}^{2}+\frac{1}{{x}^{2}}$ and ${x}^{4}+\frac{1}{{x}^{4}}.$

Let us consider the following equation:

$x-\frac{1}{x}=3$

Squaring both sides, we get:

${\left(x-\frac{1}{x}\right)}^{2}={\left(3\right)}^{2}=9\phantom{\rule{0ex}{0ex}}⇒{\left(x-\frac{1}{x}\right)}^{2}=9\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2×x×\frac{1}{x}+{\left(\frac{1}{x}\right)}^{2}=9\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2+\frac{1}{{x}^{2}}=9$
$⇒{x}^{2}+\frac{1}{{x}^{2}}=11$                               (Adding 2 to both sides)

Squaring both sides again, we get:

${\left({x}^{2}+\frac{1}{{x}^{2}}\right)}^{2}={\left(11\right)}^{2}=121\phantom{\rule{0ex}{0ex}}⇒{\left({x}^{2}+\frac{1}{{x}^{2}}\right)}^{2}=121\phantom{\rule{0ex}{0ex}}⇒{\left({x}^{2}\right)}^{2}+2\left({x}^{2}\right)\left(\frac{1}{{x}^{2}}\right)+{\left(\frac{1}{{x}^{2}}\right)}^{2}=121\phantom{\rule{0ex}{0ex}}⇒{x}^{4}+2+\frac{1}{{x}^{4}}=121\phantom{\rule{0ex}{0ex}}$
$⇒{x}^{4}+\frac{1}{{x}^{4}}=119$

#### Question 9:

If ${x}^{2}+\frac{1}{{x}^{2}}=18,$ find the values of

Let us consider the following expression:

$x+\frac{1}{x}$
Squaring the above expression, we get:

$⇒{\left(x+\frac{1}{x}\right)}^{2}=20$                                                                     ($\because$ ${x}^{2}+\frac{1}{{x}^{2}}=18$)
$⇒x+\frac{1}{x}=±\sqrt{20}$                                                            (Taking square root of both sides)

Now, let us consider the following expression:

$x-\frac{1}{x}$
Squaring the above expression, we get:

$⇒{\left(x-\frac{1}{x}\right)}^{2}=16$                                 ($\because$ ${x}^{2}+\frac{1}{{x}^{2}}=18$)
$⇒x-\frac{1}{x}=±4$                                     (Taking square root of both sides)

#### Question 10:

If x + y = 4 and xy = 2, find the value of x2 + y2

We have:

${\left(x+y\right)}^{2}={x}^{2}+2xy+{y}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}={\left(x+y\right)}^{2}-2xy$
$⇒{x}^{2}+{y}^{2}={4}^{2}-2×2$                ($\because$ )
$⇒{x}^{2}+{y}^{2}=16-4\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}=12$

#### Question 11:

If xy = 7 and xy = 9, find the value of x2 + y2

We have:

${\left(x-y\right)}^{2}={x}^{2}-2xy+{y}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}={\left(x-y\right)}^{2}+2xy$
$⇒{x}^{2}+{y}^{2}={7}^{2}+2×9$                     ($\because$ )
$⇒{x}^{2}+{y}^{2}={7}^{2}+2×9\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}=49+18\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}=67$

#### Question 12:

If 3x + 5y = 11 and xy = 2, find the value of 9x2 + 25y2

We have:

${\left(3x+5y\right)}^{2}={\left(3x\right)}^{2}+2\left(3x\right)\left(5y\right)+{\left(5y\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(3x+5y\right)}^{2}=9{x}^{2}+30xy+25{y}^{2}\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}+25{y}^{2}={\left(3x+5y\right)}^{2}-30xy$
$⇒9{x}^{2}+25{y}^{2}={11}^{2}-30×2$                          ($\because$ )
$⇒9{x}^{2}+25{y}^{2}=121-60\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}+25{y}^{2}=61$

#### Question 13:

Find the values of the following expressions:
(i) 16x2 + 24x + 9, when $x=\frac{7}{4}$
(ii) 64x2 + 81y2 + 144xy, when x = 11 and $y=\frac{4}{3}$
(iii) 81x2 + 16y2 − 72xy, when $x=\frac{2}{3}$ and $y=\frac{3}{4}$

(i) Let us consider the following expression:

$16{x}^{2}+24x+9$

Now

$16{x}^{2}+24x+9={\left(4x+3\right)}^{2}$                                 (Using identity ${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$)

(ii) Let us consider the following expression:

$64{x}^{2}+81{y}^{2}+144xy$

Now

$64{x}^{2}+81{y}^{2}+144xy={\left(8x+9y\right)}^{2}$                                (Using identity ${\left(a+b\right)}^{2}={a}^{2}+2ab+{b}^{2}$)

(iii) Let us consider the following expression:

$81{x}^{2}+16{y}^{2}-72xy$

Now

$81{x}^{2}+16{y}^{2}-72xy={\left(9x-4y\right)}^{2}$                                  (Using identity ${\left(a+b\right)}^{2}={a}^{2}-2ab+{b}^{2}$)

#### Question 14:

If $x+\frac{1}{x}=9,$ find the value of ${x}^{4}+\frac{1}{{x}^{4}}.$

Let us consider the following equation:

$x+\frac{1}{x}=9$

Squaring both sides, we get:

${\left(x+\frac{1}{x}\right)}^{2}={\left(9\right)}^{2}=81\phantom{\rule{0ex}{0ex}}⇒{\left(x+\frac{1}{x}\right)}^{2}=81\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2×x×\frac{1}{x}+{\left(\frac{1}{x}\right)}^{2}=81\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2+\frac{1}{{x}^{2}}=81$
$⇒{x}^{2}+\frac{1}{{x}^{2}}=79$                               (Subtracting 2 from both sides)

Now, squaring both sides again, we get:

${\left({x}^{2}+\frac{1}{{x}^{2}}\right)}^{2}={\left(79\right)}^{2}=6241\phantom{\rule{0ex}{0ex}}⇒{\left({x}^{2}+\frac{1}{{x}^{2}}\right)}^{2}=6241\phantom{\rule{0ex}{0ex}}⇒{\left({x}^{2}\right)}^{2}+2\left({x}^{2}\right)\left(\frac{1}{{x}^{2}}\right)+{\left(\frac{1}{{x}^{2}}\right)}^{2}=6241\phantom{\rule{0ex}{0ex}}⇒{x}^{4}+2+\frac{1}{{x}^{4}}=6241\phantom{\rule{0ex}{0ex}}$
$⇒{x}^{4}+\frac{1}{{x}^{4}}=6239$

#### Question 15:

If $x+\frac{1}{x}=12,$ find the value of $x-\frac{1}{x}.$

Let us consider the following equation:

$x+\frac{1}{x}=12$

Squaring both sides, we get:

$⇒{x}^{2}+\frac{1}{{x}^{2}}=142$                               (Subtracting 2 from both sides)

Now

#### Question 16:

If 2x + 3y = 14 and 2x − 3y = 2, find the value of xy.
[Hint: Use (2x + 3y)2 − (2x − 3y)2 = 24xy]

We will use the identity $\left(a+b\right)\left(a-b\right)={a}^{2}-{b}^{2}$ to obtain the value of xy.

${\left(2x+3y\right)}^{2}-{\left(2x-3y\right)}^{2}=\left\{\left(2x+3y\right)+\left(2x-3y\right)\right\}\left\{\left(2x+3y\right)-\left(2x-3y\right)\right\}=4x×6y=24xy\phantom{\rule{0ex}{0ex}}⇒{\left(2x+3y\right)}^{2}-{\left(2x-3y\right)}^{2}=24xy$

#### Question 17:

If x2 + y2 = 29 and xy = 2, find the value of
(i) x + y
(ii) x − y
(iii) x4 + y4

(i) We have:

(ii) We have:

(iii) We have:

#### Question 18:

What must be added to each of the following expressions to make it a whole square?
(i) 4x2 − 12x + 7
(ii) 4x2 − 20x + 20

(i) Let us consider the following expression:

$4{x}^{2}-12x+7$
The above expression can be written as:

$4{x}^{2}-12x+7={\left(2x\right)}^{2}-2×2x×3+7$
It is evident that if 2x is considered as the first term and 3 is considered as the second term, 2 is required to be added to the above expression to make it a perfect square. Therefore, 7 must become 9.

Therefore, adding and subtracting 2 in the above expression, we get:

$\left(4{x}^{2}-12x+7\right)+2-2=\left\{{\left(2x\right)}^{2}-2×2x×3+7\right\}+2-2=\left\{{\left(2x\right)}^{2}-2×2x×3+9\right\}-2={\left(2x+3\right)}^{2}-2$

(ii) Let's consider the following expression:

$4{x}^{2}-20x+20$
The above expression can be written as:

$4{x}^{2}-20x+20={\left(2x\right)}^{2}-2×2x×5+20$
It is evident that if 2x is considered as the first term and 5 is considered as the second term, 5 is required to be added to the above expression to make it a perfect square. Therefore, number 20 must become 25.

Therefore, adding and subtracting 5 in the above expression, we get:

$\left(4{x}^{2}-20x+20+5\right)-5=\left\{{\left(2x\right)}^{2}-2×2x×5+20\right\}+5-5=\left\{{\left(2x\right)}^{2}-2×2x×5+25\right\}-5={\left(2x+5\right)}^{2}-5$

#### Question 19:

Simplify:
(i) (x − y)(x + y) (x2 + y2)(x4 + y2)
(ii) (2x − 1)(2x + 1)(4x2 + 1)(16x4 + 1)
(iii) (4m − 8n)2 + (7m + 8n)2
(iv) (2.5p − 1.5q)2 − (1.5p − 2.5q)2
(v) (m2n2m)2 + 2m3n2

To simplify, we will proceed as follows:
(i)

#### Question 20:

Show that:
(i) (3x + 7)2 − 84x = (3x − 7)2
(ii) (9a − 5b)2 + 180ab = (9a + 5b)2
(iii) ${\left(\frac{4m}{3}-\frac{3n}{4}\right)}^{2}+2mn=\frac{16{m}^{2}}{9}+\frac{9{n}^{2}}{16}$
(iv) (4pq + 3q)2 − (4pq − 3q)2 = 48pq2
(v) (a − b)(a + b) + (b − c)(b + c) + (c − a)( c + a) = 0

#### Question 1:

Find the following products:
(i) (x + 4) (x + 7)
(ii) (x − 11) (x + 4)
(iii) (x + 7) (x − 5)
(iv) (x − 3) ( x − 2)
(v) (y2 − 4) (y2 − 3)
(vi) $\left(x+\frac{4}{3}\right)\left(x+\frac{3}{4}\right)$
(vii) (3x + 5) (3x + 11)
(viii) (2x2 − 3) (2x2 + 5)
(ix) (z2 + 2) (z2 − 3)
(x) (3x − 4y) (2x − 4y)
(xi) (3x2 − 4xy) (3x2 − 3xy)
(xii)
(xiii) $\left(z+\frac{3}{4}\right)\left(z+\frac{4}{3}\right)$
(xiv) (x2 + 4) (x2 + 9)
(xv) (y2 + 12) (y2 + 6)
(xvi) $\left({y}^{2}+\frac{5}{7}\right)\left({y}^{2}-\frac{14}{5}\right)$
(xvii) (p2 + 16) $\left({p}^{2}-\frac{1}{4}\right)$

(i) Here, we will use the identity $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab.$
$\left(x+4\right)\left(x+7\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+\left(4+7\right)x+4×7\phantom{\rule{0ex}{0ex}}={x}^{2}+11x+28$

(ii) Here, we will use the identity $\left(x-a\right)\left(x+b\right)={x}^{2}+\left(b-a\right)x-ab$.
$\left(x-11\right)\left(x+4\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+\left(4-11\right)x-11×4\phantom{\rule{0ex}{0ex}}={x}^{2}-7x-44$

(iii) Here, we will use the identity​ $\left(x+a\right)\left(x-b\right)={x}^{2}+\left(a-b\right)x-ab$.
$\left(x+7\right)\left(x-5\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+\left(7-5\right)x-7×5\phantom{\rule{0ex}{0ex}}={x}^{2}+2x-35$

(iv) Here, we will use the identity​ $\left(x-a\right)\left(x-b\right)={x}^{2}-\left(a+b\right)x+ab$.
$\left(x-3\right)\left(x-2\right)\phantom{\rule{0ex}{0ex}}={x}^{2}-\left(3+2\right)x+3×2\phantom{\rule{0ex}{0ex}}={x}^{2}-5x+6$

(v) Here, we will use the identity​ $\left(x-a\right)\left(x-b\right)={x}^{2}-\left(a+b\right)x+ab$.
$\left({y}^{2}-4\right)\left({y}^{2}-3\right)\phantom{\rule{0ex}{0ex}}={\left({y}^{2}\right)}^{2}-\left(4+3\right)\left({y}^{2}\right)+4×3\phantom{\rule{0ex}{0ex}}={y}^{4}-7{y}^{2}+12$

(vi) Here, we will use the identity​ $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab$.
$\left(x+\frac{4}{3}\right)\left(x+\frac{3}{4}\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+\left(\frac{4}{3}+\frac{3}{4}\right)x+\frac{4}{3}×\frac{3}{4}\phantom{\rule{0ex}{0ex}}={x}^{2}+\frac{25}{12}x+1$

(vii) Here, we will use the identity​ $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab$.
$\left(3x+5\right)\left(3x+11\right)\phantom{\rule{0ex}{0ex}}={\left(3x\right)}^{2}+\left(5+11\right)\left(3x\right)+5×11\phantom{\rule{0ex}{0ex}}=9{x}^{2}+48x+55$

(viii) Here, we will use the identity​ $\left(x-a\right)\left(x+b\right)={x}^{2}+\left(b-a\right)x-ab$.
$\left(2{x}^{2}-3\right)\left(2{x}^{2}+5\right)\phantom{\rule{0ex}{0ex}}={\left(2{x}^{2}\right)}^{2}+\left(5-3\right)\left(2{x}^{2}\right)-3×5\phantom{\rule{0ex}{0ex}}=4{x}^{4}+4{x}^{2}-15$

(ix) Here, we will use the identity​ $\left(x+a\right)\left(x-b\right)={x}^{2}+\left(a-b\right)x-ab$.
$\left({z}^{2}+2\right)\left({z}^{2}-3\right)\phantom{\rule{0ex}{0ex}}={\left({z}^{2}\right)}^{2}+\left(2-3\right)\left({z}^{2}\right)-2×3\phantom{\rule{0ex}{0ex}}={z}^{4}-{z}^{2}-6$

(x) Here, we will use the identity​ $\left(x-a\right)\left(x-b\right)={x}^{2}-\left(a+b\right)x+ab$.

(xi) Here, we will use the identity​ $\left(x-a\right)\left(x-b\right)={x}^{2}-\left(a+b\right)x+ab$.

$\left(3{x}^{2}-4xy\right)\left(3{x}^{2}-3xy\right)\phantom{\rule{0ex}{0ex}}={\left(3{x}^{2}\right)}^{2}-\left(4xy+3xy\right)\left(3{x}^{2}\right)+4xy×3xy\phantom{\rule{0ex}{0ex}}=9{x}^{4}-\left(12{x}^{3}y+9{x}^{3}y\right)+12{x}^{2}{y}^{2}\phantom{\rule{0ex}{0ex}}=9{x}^{4}-21{x}^{3}y+12{x}^{2}{y}^{2}$

(xii) Here, we will use the identity​ $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab$.
$\left(x+\frac{1}{5}\right)\left(x+5\right)\phantom{\rule{0ex}{0ex}}={x}^{2}+\left(\frac{1}{5}+5\right)x+\frac{1}{5}×5\phantom{\rule{0ex}{0ex}}={x}^{2}+\frac{26}{5}x+1$

(xiii) Here, we will use the identity​ $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab$.
$\left(z+\frac{3}{4}\right)\left(z+\frac{4}{3}\right)\phantom{\rule{0ex}{0ex}}={z}^{2}+\left(\frac{3}{4}+\frac{4}{3}\right)x+\frac{3}{4}×\frac{4}{3}\phantom{\rule{0ex}{0ex}}={z}^{2}+\frac{25}{12}z+1$

(xiv) Here, we will use the identity​ $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab$.
$\left({x}^{2}+4\right)\left({x}^{2}+9\right)\phantom{\rule{0ex}{0ex}}={\left({x}^{2}\right)}^{2}+\left(4+9\right)\left({x}^{2}\right)+4×9\phantom{\rule{0ex}{0ex}}={x}^{4}+13{x}^{2}+36$

(xv) Here, we will use the identity​ $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab$.
$\left({y}^{2}+12\right)\left({y}^{2}+6\right)\phantom{\rule{0ex}{0ex}}={\left({y}^{2}\right)}^{2}+\left(12+6\right)\left({y}^{2}\right)+12×6\phantom{\rule{0ex}{0ex}}={y}^{4}+18{y}^{2}+72$

(xvi) Here, we will use the identity​ $\left(x+a\right)\left(x-b\right)={x}^{2}+\left(a-b\right)x-ab$.
$\left({y}^{2}+\frac{5}{7}\right)\left({y}^{2}-\frac{14}{5}\right)\phantom{\rule{0ex}{0ex}}={\left({y}^{2}\right)}^{2}+\left(\frac{5}{7}-\frac{14}{5}\right)\left({y}^{2}\right)-\frac{5}{7}×\frac{14}{5}\phantom{\rule{0ex}{0ex}}={y}^{4}-\frac{73}{35}{y}^{2}-2$

(xvii) Here, we will use the identity​ $\left(x+a\right)\left(x-b\right)={x}^{2}+\left(a-b\right)x-ab$.
$\left({p}^{2}+16\right)\left({p}^{2}-\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}={\left({p}^{2}\right)}^{2}+\left(16-\frac{1}{4}\right)\left({p}^{2}\right)-16×\frac{1}{4}\phantom{\rule{0ex}{0ex}}={p}^{4}+\frac{63}{4}{p}^{2}-4$

#### Question 2:

Evaluate the following:
(i) 102 × 106
(ii) 109 × 107
(iii) 35 × 37
(iv) 53 × 55
(v) 103 × 96
(vi) 34 × 36
(vii) 994 × 1006

(i) Here, we will use the identity $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab$
$102×106\phantom{\rule{0ex}{0ex}}=\left(100+2\right)\left(100+6\right)\phantom{\rule{0ex}{0ex}}={100}^{2}+\left(2+6\right)100+2×6\phantom{\rule{0ex}{0ex}}=10000+800+12\phantom{\rule{0ex}{0ex}}=10812$

(ii) Here, we will use the identity $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab$
$109 × 107\phantom{\rule{0ex}{0ex}}=\left(100+9\right)\left(100+7\right)\phantom{\rule{0ex}{0ex}}={100}^{2}+\left(9+7\right)100+9×7\phantom{\rule{0ex}{0ex}}=10000+1600+63\phantom{\rule{0ex}{0ex}}=11663$

(iii) Here, we will use the identity $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab$
$35 × 37\phantom{\rule{0ex}{0ex}}=\left(30+5\right)\left(30+7\right)\phantom{\rule{0ex}{0ex}}={30}^{2}+\left(5+7\right)30+5×7\phantom{\rule{0ex}{0ex}}=900+360+35\phantom{\rule{0ex}{0ex}}=1295$

(iv) Here, we will use the identity $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab$
$53 × 55\phantom{\rule{0ex}{0ex}}=\left(50+3\right)\left(50+5\right)\phantom{\rule{0ex}{0ex}}={50}^{2}+\left(3+5\right)50+3×5\phantom{\rule{0ex}{0ex}}=2500+400+15\phantom{\rule{0ex}{0ex}}=2915$

(v) Here, we will use the identity $\left(x+a\right)\left(x-b\right)={x}^{2}+\left(a-b\right)x-ab$
$103 × 96\phantom{\rule{0ex}{0ex}}=\left(100+3\right)\left(100-4\right)\phantom{\rule{0ex}{0ex}}={100}^{2}+\left(3-4\right)100-3×4\phantom{\rule{0ex}{0ex}}=10000-100-12\phantom{\rule{0ex}{0ex}}=9888$

(vi) Here, we will use the identity $\left(x+a\right)\left(x+b\right)={x}^{2}+\left(a+b\right)x+ab$
$34 × 36\phantom{\rule{0ex}{0ex}}=\left(30+4\right)\left(30+6\right)\phantom{\rule{0ex}{0ex}}={30}^{2}+\left(4+6\right)30+4×6\phantom{\rule{0ex}{0ex}}=900+300+24\phantom{\rule{0ex}{0ex}}=1224$

(vii) Here, we will use the identity $\left(x-a\right)\left(x+b\right)={x}^{2}+\left(b-a\right)x-ab$
$994 × 1006\phantom{\rule{0ex}{0ex}}=\left(1000-6\right)×\left(1000+6\right)\phantom{\rule{0ex}{0ex}}={1000}^{2}+\left(6-6\right)×1000-6×6\phantom{\rule{0ex}{0ex}}=1000000-36\phantom{\rule{0ex}{0ex}}=999964$

#### Question 1:

(i) 3a2b, − 4a2b, 9a2b
(ii)
(iii) 4xy2 − 7x2y, 12x2y − 6xy2, − 3x2y +5xy2
(iv)
(v)
(vi)

(i) To add the like terms, we proceed as follows:

(ii) To add the like terms, we proceed as follows:

(iii) To add, we proceed as follows:

(iv) To add, we proceed as follows:

(v) To add, we proceed as follows:

(vi) To add, we proceed as follows:

#### Question 2:

Subtract:
(i) − 5xy from 12xy
(ii) 2a2 from − 7a2
(iii) 2a − b from 3a − 5b
(iv) 2x3 − 4x2 + 3x + 5 from 4x3 + x2 + x + 6
(v)
(vi)
(vii)
(viii)

#### Question 3:

Take away:
(i)
(ii)
(iii)
(iv)
(v)

(i) The difference is given by:

$\left(\frac{{x}^{3}}{3}-\frac{5}{2}{x}^{2}+\frac{3}{5}x+\frac{1}{4}\right)-\left(\frac{6}{5}{x}^{2}-\frac{4}{5}{x}^{3}+\frac{5}{6}+\frac{3}{2}x\right)\phantom{\rule{0ex}{0ex}}=\frac{{x}^{3}}{3}-\frac{5}{2}{x}^{2}+\frac{3}{5}x+\frac{1}{4}-\frac{6}{5}{x}^{2}+\frac{4}{5}{x}^{3}-\frac{5}{6}-\frac{3}{2}x$
$=\frac{{x}^{3}}{3}+\frac{4}{5}{x}^{3}-\frac{5}{2}{x}^{2}-\frac{6}{5}{x}^{2}+\frac{3}{5}x-\frac{3}{2}x+\frac{1}{4}-\frac{5}{6}$             (Collecting like terms)
=$\left(\frac{5+12}{15}\right){x}^{3}+\left(\frac{-25-12}{10}\right){x}^{2}+\left(\frac{6-15}{10}\right)x+\left(\frac{6-20}{24}\right)$
$=\frac{17}{15}{x}^{3}-\frac{37}{10}{x}^{2}-\frac{9}{10}x-\frac{7}{12}$                                            (Combining like terms)

(ii) The difference is given by:

$\left(\frac{1}{3}{a}^{3}-\frac{3{a}^{2}}{4}-\frac{5}{2}\right)-\left(\frac{5{a}^{2}}{2}+\frac{3{a}^{3}}{2}+\frac{a}{3}-\frac{6}{5}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}{a}^{3}-\frac{3{a}^{2}}{4}-\frac{5}{2}-\frac{5{a}^{2}}{2}-\frac{3{a}^{3}}{2}-\frac{a}{3}+\frac{6}{5}$
$=\frac{1}{3}{a}^{3}-\frac{3{a}^{3}}{2}-\frac{3{a}^{2}}{4}-\frac{5{a}^{2}}{2}-\frac{a}{3}-\frac{5}{2}+\frac{6}{5}$                      (Collecting like terms)
=$\left(\frac{2-9}{6}\right){a}^{3}+\left(\frac{-3-10}{4}\right){a}^{2}-\frac{a}{3}+\left(\frac{-25+12}{10}\right)$
$=-\frac{7}{6}{a}^{3}-\frac{13}{4}{a}^{2}-\frac{a}{3}-\frac{13}{10}$                                             (Combining like terms)

(iii) The difference is given by:

$\left(\frac{7}{2}-\frac{x}{3}-\frac{{x}^{2}}{5}\right)-\left(\frac{7{x}^{3}}{4}+\frac{3{x}^{2}}{5}+\frac{x}{2}+\frac{9}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{7}{2}-\frac{x}{3}-\frac{{x}^{2}}{5}-\frac{7{x}^{3}}{4}-\frac{3{x}^{2}}{5}-\frac{x}{2}-\frac{9}{2}$
$=\frac{7}{2}-\frac{9}{2}-\frac{x}{3}-\frac{x}{2}-\frac{{x}^{2}}{5}-\frac{3{x}^{2}}{5}-\frac{7{x}^{3}}{4}$                     (Collecting like terms)
=$\left(\frac{7-9}{2}\right)+\left(\frac{-2-3}{6}\right)x+\left(\frac{-1-3}{5}\right){x}^{2}-\frac{7{x}^{3}}{4}$
$=-1-\frac{5x}{6}-\frac{4{x}^{2}}{5}-\frac{7{x}^{3}}{4}$                                           (Combining like terms )

(iv) The difference is given by:

$\left(\frac{1}{3}-\frac{5}{3}{y}^{2}\right)-\left(\frac{{y}^{3}}{3}+\frac{7{y}^{2}}{3}+\frac{y}{2}+\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}-\frac{5}{3}{y}^{2}-\frac{{y}^{3}}{3}-\frac{7{y}^{2}}{3}-\frac{y}{2}-\frac{1}{2}$
$=\frac{1}{3}-\frac{1}{2}-\frac{y}{2}-\frac{5}{3}{y}^{2}-\frac{7{y}^{2}}{3}-\frac{{y}^{3}}{3}$             ( Collecting like terms)
=$\left(\frac{2-3}{6}\right)-\frac{y}{2}+\left(\frac{-5-7}{3}\right){y}^{2}-\frac{{y}^{3}}{3}$
$=-\frac{1}{6}-\frac{y}{2}-4{y}^{2}-\frac{{y}^{3}}{3}$                               (Combining like terms. )

(v) The difference is given by:

$\left(\frac{3}{2}ab-\frac{7}{4}ac-\frac{5}{6}bc\right)-\left(\frac{2}{3}ac-\frac{5}{7}ab+\frac{2}{3}bc\right)\phantom{\rule{0ex}{0ex}}=\frac{3}{2}ab-\frac{7}{4}ac-\frac{5}{6}bc-\frac{2}{3}ac+\frac{5}{7}ab-\frac{2}{3}bc$

$=\frac{3}{2}ab+\frac{5}{7}ab-\frac{7}{4}ac-\frac{2}{3}ac-\frac{5}{6}bc-\frac{2}{3}bc$             (Collecting like terms )
=$\left(\frac{21+10}{14}\right)ab+\left(\frac{-21-8}{12}\right)ac+\left(\frac{-5-4}{6}\right)bc$
$=\frac{31}{14}ab-\frac{29}{12}ac-\frac{3}{2}bc$                                         (Combining like terms )

#### Question 4:

Subtract 3x − 4y − 7z from the sum of x − 3y + 2z and − 4x + 9y − 11z.

Let first add the expressions . We get:

$\left(x-3y+2z\right)+\left(-4x+9y-11z\right)$
$=x-3y+2z-4x+9y-11z$
$=x-4x-3y+9y+2z-11z$                     (Collecting like terms)
$=-3x+6y-9z$                                       (Combining like terms)

Now, subtracting the expression $3x-4y-7z$ from the above sum; we get:

$\left(-3x+6y-9z\right)-\left(3x-4y-7z\right)\phantom{\rule{0ex}{0ex}}=-3x+6y-9z-3x+4y+7z$
$=-3x-3x+6y+4y-9z+7z$          (Collecting like terms)
$=-6x+10y-2z$                              (Combining like terms)

Thus, the answer is $-6x+10y-2z$.

#### Question 5:

Subtract the sum of 3l − 4m − 7n2 and 2l + 3m − 4n2 from the sum of 9l + 2m − 3n2 and − 3l + m + 4n2 .....

We have to subtract the sum of (3l $-$ 4m $-$ 7n2) and (2l + 3m $-$ 4n2) from the sum of (9l + 2m $-$ 3n2) and  ($-$3l m + 4n2)​

$\left\{\left(9l+2m-3{n}^{2}\right)+\left(-3l+m+4{n}^{2}\right)\right\}-\left\{\left(3l-4m-7{n}^{2}\right)+\left(2l+3m-4{n}^{2}\right)\right\}$
$=\left(9l-3l+2m+m-3{n}^{2}+4{n}^{2}\right)-\left(3l+2l-4m+3m-7{n}^{2}-4{n}^{2}\right)$
(Combining like terms inside the parentheses)
$=6l+3m+{n}^{2}-5l+m+11{n}^{2}$
$=6l-5l+3m+m+{n}^{2}+11{n}^{2}$                                                        (Collecting like terms)
$=l+4m+12{n}^{2}$                                                                              (Combining like terms)

Thus, the required solution is $l+4m+12{n}^{2}$.

#### Question 6:

Subtract the sum of 2xx2 + 5 and − 4x − 3 + 7x2 from 5.

We have to subtract the sum of (2x  x2 + 5) and (4x − 3 + 7x) from 5.

$5-\left\{\left(2x-{x}^{2}+5\right)+\left(-4x-3+7{x}^{2}\right)\right\}\phantom{\rule{0ex}{0ex}}=5-\left(2x-4x-{x}^{2}+7{x}^{2}+5-3\right)\phantom{\rule{0ex}{0ex}}=5-2x+4x+{x}^{2}-7{x}^{2}-5+3$
$=5-5+3-2x+4x+{x}^{2}-7{x}^{2}$                (Collecting like terms)
$=3+2x-6{x}^{2}$                                        (Combining like terms)

Thus, the answer is $3+2x-6{x}^{2}$.

#### Question 7:

Simplify each of the following:
(i) x2 − 3x + 5 − $\frac{1}{2}$ (3x2 − 5x + 7)
(ii) [5 − 3x + 2y − (2xy)] − (3x − 7y + 9)
(iii) $\frac{11}{2}{x}^{2}y-\frac{9}{4}x{y}^{2}+\frac{1}{4}xy-\frac{1}{14}{y}^{2}x+\frac{1}{15}y{x}^{2}+\frac{1}{2}xy$
(iv) $\left(\frac{1}{3}{y}^{2}-\frac{4}{7}y+11\right)-\left(\frac{1}{7}y-3+2{y}^{2}\right)-\left(\frac{2}{7}y-\frac{2}{3}{y}^{2}+2\right)$
(v) $-\frac{1}{2}{a}^{2}{b}^{2}c+\frac{1}{3}a{b}^{2}c-\frac{1}{4}ab{c}^{2}-\frac{1}{5}c{b}^{2}{a}^{2}+\frac{1}{6}c{b}^{2}a-\frac{1}{7}{c}^{2}ab+\frac{1}{8}c{a}^{2}b.$

Thus, the answer is $-\frac{{x}^{2}}{2}-\frac{x}{2}+\frac{3}{2}$.

$=\frac{11}{2}{x}^{2}y+\frac{1}{15}y{x}^{2}-\frac{9}{4}x{y}^{2}-\frac{1}{14}{y}^{2}x+\frac{1}{4}xy+\frac{1}{2}xy$           (Collecting like terms)
=$\left(\frac{165+2}{30}\right){x}^{2}y+\left(\frac{-63-2}{28}\right)x{y}^{2}+\left(\frac{1+2}{4}\right)xy$
$=\frac{167}{30}{x}^{2}y-\frac{65}{28}{y}^{2}x+\frac{3}{2}xy$                                                (Combining like terms)

$=\frac{1}{3}{y}^{2}-2{y}^{2}+\frac{2}{3}{y}^{2}-\frac{4}{7}y-\frac{1}{7}y-\frac{2}{7}y+11+3-2$        (Collecting like terms)
=$\left(\frac{1-6+2}{3}\right){y}^{2}+\left(\frac{-4-1-2}{7}\right)y+12$
$=-{y}^{2}-7y+12$                                                             (Combining like terms)

$=-\frac{1}{2}{a}^{2}{b}^{2}c-\frac{1}{5}c{b}^{2}{a}^{2}+\frac{1}{3}a{b}^{2}c+\frac{1}{6}c{b}^{2}a-\frac{1}{4}ab{c}^{2}-\frac{1}{7}{c}^{2}ab+\frac{1}{8}c{a}^{2}b$           (Collecting like terms)
=
$=-\frac{7}{10}{a}^{2}{b}^{2}c+\frac{1}{2}a{b}^{2}c-\frac{11}{28}ab{c}^{2}+\frac{1}{8}{a}^{2}bc$                                                    (Combining like terms)

View NCERT Solutions for all chapters of Class 8