Rs Aggarwal 2018 Solutions for Class 8 Math Chapter 6 Operations On Algebraic Expressions are provided here with simple step-by-step explanations. These solutions for Operations On Algebraic Expressions are extremely popular among Class 8 students for Math Operations On Algebraic Expressions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 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 Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### Question 1:

8ab, −5ab, 3ab, −ab

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

________

$5ab$

#### Question 2:

7x, −3x, 5x, −x, −2x

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

_____

#### Question 3:

3a − 4b + 4c, 2a + 3b − 8c, a − 6b + c

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

___________

#### Question 4:

5x − 8y + 2z, 3z − 4y − 2x, 6yzx and 3x − 2z − 3y

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

#### Question 5:

6ax − 2by + 3cz, 6by − 11axcz and 10cz − 2ax − 3by

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

#### Question 6:

2x3 − 9x2 + 8, 3x2 − 6x − 5, 7x3 − 10x + 1 and 3 + 2x − 5x2 − 4x3

On arranging the terms of the given expressions in the descending powers of $x$ and adding column-wise:

#### Question 7:

6p + 4q r + 3, 2r − 5p − 6, 11q − 7p + 2r − 1 and 2q − 3r + 4

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise:

#### Question 8:

4x2 − 7xy + 4y2 − 3, 5 + 6y2 − 8xy + x2 and 6 − 2xy + 2x2 − 5y2

On arranging the terms of the given expressions in the descending powers of $x$ and adding column-wise:

#### Question 9:

Subtract:
3a2b from −5a2b

On arranging the terms of the given expressions in the descending powers of $x$ and subtracting:

#### Question 10:

Subtract:
−8pq from 6pq

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

#### Question 11:

Subtract:
−2abc from −8abc

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

#### Question 12:

Subtract:
−16p from −11p

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

#### Question 13:

Subtract:
2a − 5b + 2c − 9 from 3a − 4b c + 6

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

#### Question 14:

Subtract:
−6p + q + 3r + 8 from p − 2q − 5r − 8

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

#### Question 15:

Subtract:
x3 + 3x2 − 5x + 4 from 3x3x2 + 2x − 4

On arranging the terms of the given expressions in the descending powers of $x$ and subtracting column-wise:

#### Question 16:

Subtract:
5y4 − 3y3 + 2y2 + y − 1 from 4y4 − 2y3 − 6y2y + 5

Arranging the terms of the given expressions in the descending powers of $x$ and subtracting column-wise:

#### Question 17:

Subtract:
4p2 + 5q2 − 6r2 + 7 from 3p2 − 4q2 − 5r2 − 6

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

#### Question 18:

What must be subtracted from 3a2 − 6ab − 3b2 − 1 to get 4a2 − 7ab − 4b2 + 1?

Let the required number be $x$.
$\left(3{a}^{2}-6ab-3{b}^{2}-1\right)-x=4{a}^{2}-7ab-4{b}^{2}+1$
$\left(3{a}^{2}-6ab-3{b}^{2}-1\right)-\left(4{a}^{2}-7ab-4{b}^{2}+1\right)=x\phantom{\rule{0ex}{0ex}}$

∴ Required number = $-{a}^{2}+ab+{b}^{2}-2$

#### Question 19:

The two adjacent sides of a rectangle are 5x2 − 3y2 and x2 + 2xy. Find the perimeter.

Sides of the rectangle are $l$ and $b$.
$l=5{x}^{2}-3{y}^{2}\phantom{\rule{0ex}{0ex}}b={x}^{2}+2xy$
Perimeter of the rectangle is $\left(2l+2b\right)$.

#### Question 20:

The perimeter of a triangle is 6p2 − 4p + 9 and two of its sides are p2 − 2p + 1 and 3p2 − 5p + 3. Find the third side of the triangle.

Let be the three sides of the triangle.

∴ Perimeter of the triangle =$\left(a+b+c\right)$

Given perimeter of the triangle = $6{p}^{2}-4p+9$
One side ($a$)  = ${p}^{2}-2p+1$
Other side ($b$) = $3{p}^{2}-5p+3$
Perimeter = $\left(a+b+c\right)$

Thus, the third side is $2{p}^{2}+3p+5$.

#### Question 1:

Find the product:
(5x + 7) × (3x + 4)

By horizontal method:
$\left(5x+7\right)×\left(3x+4\right)\phantom{\rule{0ex}{0ex}}=5x\left(3x+4\right)+7\left(3x+4\right)\phantom{\rule{0ex}{0ex}}=15{x}^{2}+20x+21x+28\phantom{\rule{0ex}{0ex}}=15{x}^{2}+41x+28$

#### Question 2:

Find the product:
(4x + 9) × (x − 6)

By horizontal method:

$\left(4x+9\right)×\left(x-6\right)\phantom{\rule{0ex}{0ex}}=4x\left(x-6\right)+9\left(x-6\right)\phantom{\rule{0ex}{0ex}}=4{x}^{2}-24x+9x-54\phantom{\rule{0ex}{0ex}}=4{x}^{2}-15x-54$

#### Question 3:

Find the product:
(2x + 5) × (4x − 3)

By horizontal method:

$\left(2x+5\right)×\left(4x-3\right)\phantom{\rule{0ex}{0ex}}=2x\left(4x-3\right)+5\left(4x-3\right)\phantom{\rule{0ex}{0ex}}=8{x}^{2}-6x+20x-15\phantom{\rule{0ex}{0ex}}=8{x}^{2}+14x-15$

#### Question 4:

Find the product:
(3y − 8) × (5y − 1)

By horizontal method:

$\left(3y-8\right)×\left(5y-1\right)\phantom{\rule{0ex}{0ex}}=3y\left(5y-1\right)-8\left(5y-1\right)\phantom{\rule{0ex}{0ex}}=15{y}^{2}-3y-40y+8\phantom{\rule{0ex}{0ex}}=15{y}^{2}-43y+8$

#### Question 5:

Find the product:
(7x + 2y) × (x + 4y)

By horizontal method:

$\left(7x+2y\right)×\left(x+4y\right)\phantom{\rule{0ex}{0ex}}=7x\left(x+4y\right)+2y\left(x+4y\right)\phantom{\rule{0ex}{0ex}}=7{x}^{2}+28xy+2xy+8{y}^{2}\phantom{\rule{0ex}{0ex}}=7{x}^{2}+30xy+8{y}^{2}$

#### Question 6:

Find the product:
(9x + 5y) × (4x + 3y)

By horizontal method:

$\left(9x+5y\right)×\left(4x+3y\right)\phantom{\rule{0ex}{0ex}}9x\left(4x+3y\right)+5y\left(4x+3y\right)\phantom{\rule{0ex}{0ex}}=36{x}^{2}+27xy+20xy+15{y}^{2}\phantom{\rule{0ex}{0ex}}=36{x}^{2}+47xy+15{y}^{2}$

#### Question 7:

Find the product:
(3m − 4n) × (2m − 3n)

By horizontal method:

$\left(3m-4n\right)×\left(2m-3n\right)\phantom{\rule{0ex}{0ex}}=3m\left(2m-3n\right)-4n\left(2m-3n\right)\phantom{\rule{0ex}{0ex}}=6{m}^{2}-9mn-8mn+12{n}^{2}\phantom{\rule{0ex}{0ex}}=6{m}^{2}-17mn+12{n}^{2}$

#### Question 8:

Find the product:
(x2a2) × (xa)

By horizontal method:

$\left({x}^{2}-{a}^{2}\right)×\left(x-a\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(x-a\right)-{a}^{2}\left(x-a\right)\phantom{\rule{0ex}{0ex}}={x}^{3}-a{x}^{2}-{a}^{2}x+{a}^{3}$
i.e $\left({x}^{3}+{a}^{3}\right)-ax\left(x-a\right)$

#### Question 9:

Find the product:
(x2y2) × (x + 2y)

By horizontal method:

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

#### Question 10:

Find the product:
(3p2 + q2) × (2p2 − 3q2)

By horizontal method:

$\left(3{p}^{2}+{q}^{2}\right)×\left(2{p}^{2}-3{q}^{2}\right)\phantom{\rule{0ex}{0ex}}=3{p}^{2}\left(2{p}^{2}-3{q}^{2}\right)+{q}^{2}\left(2{p}^{2}-3{q}^{2}\right)\phantom{\rule{0ex}{0ex}}=6{p}^{4}-9{p}^{2}{q}^{2}+2{p}^{2}{q}^{2}-3{q}^{4}\phantom{\rule{0ex}{0ex}}i.e6{p}^{4}-7{p}^{2}{q}^{2}-3{q}^{4}$

#### Question 11:

Find the product:
(2x2 − 5y2) × (x2 + 3y2)

By horizontal method:

$\left(2{x}^{2}-5{y}^{2}\right)×\left({x}^{2}+3{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=2{x}^{2}\left({x}^{2}+3{y}^{2}\right)-5{y}^{2}\left({x}^{2}+3{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=2{x}^{4}+6{x}^{2}{y}^{2}-5{x}^{2}{y}^{2}-15{y}^{4}\phantom{\rule{0ex}{0ex}}=2{x}^{4}+{x}^{2}{y}^{2}-15{y}^{4}\phantom{\rule{0ex}{0ex}}$

#### Question 12:

Find the product:
(x3y3) × (x2 + y2)

By horizontal method:

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

#### Question 13:

Find the product:
(x4 + y4) × (x2y2)

By horizontal method:
$\left({x}^{4}+{y}^{4}\right)×\left({x}^{2}-{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{4}\left({x}^{2}-{y}^{2}\right)+{y}^{4}\left({x}^{2}-{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{6}-{x}^{4}{y}^{2}+{y}^{4}{x}^{2}-{y}^{6}\phantom{\rule{0ex}{0ex}}=\left({x}^{6}-{y}^{6}\right)-{x}^{2}{y}^{2}\left({x}^{2}-{y}^{2}\right)$

#### Question 14:

Find the product:
$\left({x}^{4}+\frac{1}{{x}^{4}}\right)×\left(x+\frac{1}{x}\right)$

By horizontal method:

#### Question 15:

Find the product:
(x2 − 3x + 7) × (2x + 3)

By horizontal method:

$\left({x}^{2}-3x+7\right)×\left(2x+3\right)\phantom{\rule{0ex}{0ex}}=2x\left({x}^{2}-3x+7\right)+3\left({x}^{2}-3x+7\right)\phantom{\rule{0ex}{0ex}}=2{x}^{3}-6{x}^{2}+14x+3{x}^{2}-9x+21\phantom{\rule{0ex}{0ex}}=2{x}^{3}-3{x}^{2}+5x+21\phantom{\rule{0ex}{0ex}}$

#### Question 16:

Find the product:
(3x2 + 5x − 9) × (3x − 5)

By horizontal method:
$\left(3{x}^{2}+5x-9\right)×\left(3x-5\right)\phantom{\rule{0ex}{0ex}}=3x\left(3{x}^{2}+5x-9\right)-5\left(3{x}^{2}+5x-9\right)\phantom{\rule{0ex}{0ex}}=9{x}^{3}+15{x}^{2}-27x-15{x}^{2}-25x+45\phantom{\rule{0ex}{0ex}}=9{x}^{3}-52x+45$

#### Question 17:

Find the product:
(x2xy + y2) × (x + y)

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

#### Question 18:

Find the product:
(x2 + xy + y2) × (xy)

By horizontal method:

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

#### Question 19:

Find the product:
(x3 − 2x2 + 5) × (4x − 1)

By horizontal method:

$\left({x}^{3}-2{x}^{2}+5\right)×\left(4x-1\right)\phantom{\rule{0ex}{0ex}}=4x\left({x}^{3}-2{x}^{2}+5\right)-1\left({x}^{3}-2{x}^{2}+5\right)\phantom{\rule{0ex}{0ex}}=4{x}^{4}-8{x}^{3}+20x-{x}^{3}+2{x}^{2}-5\phantom{\rule{0ex}{0ex}}=4{x}^{4}-9{x}^{3}+2{x}^{2}+20x-5$

#### Question 20:

Find the product:
(9x2x + 15) × (x2 − 3)

By horizontal method:

$\left(9{x}^{2}-x+15\right)×\left({x}^{2}-3\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(9{x}^{2}-x+15\right)-3\left(9{x}^{2}-x+15\right)\phantom{\rule{0ex}{0ex}}=9{x}^{4}-{x}^{3}+15{x}^{2}-27{x}^{2}+3x-45\phantom{\rule{0ex}{0ex}}=9{x}^{4}-{x}^{3}-12{x}^{2}+3x-45$

#### Question 21:

Find the product:
(x2 − 5x + 8) × (x2 + 2)

By horizontal method:

$\left({x}^{2}-5x+8\right)×\left({x}^{2}+2\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left({x}^{2}-5x+8\right)+2\left({x}^{2}-5x+8\right)\phantom{\rule{0ex}{0ex}}={x}^{4}-5{x}^{3}+8{x}^{2}+2{x}^{2}-10x+16\phantom{\rule{0ex}{0ex}}={x}^{4}-5{x}^{3}+10{x}^{2}-10x+16$

#### Question 22:

Find the product:
(x3 − 5x2 + 3x + 1) × (x3 − 3)

By horizontal method:

$\left({x}^{3}-5{x}^{2}+3x+1\right)×\left({x}^{2}-3\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left({x}^{3}-5{x}^{2}+3x+1\right)-3\left({x}^{3}-5{x}^{2}+3x+1\right)\phantom{\rule{0ex}{0ex}}={x}^{5}-5{x}^{4}+3{x}^{3}+{x}^{2}-3{x}^{3}+15{x}^{2}-9x-3\phantom{\rule{0ex}{0ex}}={x}^{5}-5{x}^{4}+16{x}^{2}-9x-3$

#### Question 23:

Find the product:
(3x + 2y − 4) × (xy + 2)

By horizontal method:

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

#### Question 24:

Find the product:
(x2 − 5x + 8) × (x2 + 2x − 3)

By horizontal method:

$\left({x}^{2}-5x+8\right)×\left({x}^{2}+2x-3\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left({x}^{2}-5x+8\right)+2x\left({x}^{2}-5x+8\right)-3\left({x}^{2}-5x+8\right)\phantom{\rule{0ex}{0ex}}={x}^{4}-5{x}^{3}+8{x}^{2}+2{x}^{3}-10{x}^{2}+16x-3{x}^{2}+15x-24\phantom{\rule{0ex}{0ex}}={x}^{4}-3{x}^{3}-5{x}^{2}+31x-24$

#### Question 25:

Find the product:
(2x2 + 3x − 7) × (3x2 − 5x + 4)

By horizontal method:

$\left(2{x}^{2}+3x-7\right)×\left(3{x}^{2}-5x+4\right)\phantom{\rule{0ex}{0ex}}=2{x}^{2}\left(3{x}^{2}-5x+4\right)+3x\left(3{x}^{2}-5x+4\right)-7\left(3{x}^{2}-5x+4\right)\phantom{\rule{0ex}{0ex}}=6{x}^{4}-10{x}^{3}+8{x}^{2}+9{x}^{3}-15{x}^{2}+12x-21{x}^{2}+35x-28\phantom{\rule{0ex}{0ex}}=6{x}^{4}-{x}^{3}-28{x}^{2}+47x-28$

#### Question 26:

Find the product:
(9x2x + 15) × (x2x − 1)

By horizontal method:

$\left(9{x}^{2}-x+15\right)×\left({x}^{2}-x-1\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(9{x}^{2}-x+15\right)-x\left(9{x}^{2}-x+15\right)-1\left(9{x}^{2}-x+15\right)\phantom{\rule{0ex}{0ex}}=9{x}^{4}-{x}^{3}+15{x}^{2}-9{x}^{3}+{x}^{2}-15x-9{x}^{2}+x-15\phantom{\rule{0ex}{0ex}}=9{x}^{4}-10{x}^{3}+7{x}^{2}-14x-15$

#### Question 1:

Divide:
(i) 24x2y3 by 3xy
(ii) 36xyz2 by −9xz
(iii) −72x2y2z by −12xyz
(iv) −56mnp2 by 7mnp

(i) 24x2y3 by 3xy

Therefore, the quotient is 8xy2.

(ii) 36xyz2 by −9xz

Therefore, the quotient is 4yz.

(iii)

Therefore, the quotient is 6xy.

(iv) −56mnp2 by 7mnp

Therefore, the quotient is −8p.

#### Question 2:

Divide:
(i) 5m3 − 30m2 + 45m by 5m
(ii) 8x2y2 − 6xy2 + 10x2y3 by 2xy
(iii) 9x2y − 6xy + 12xy2 by − 3xy
(iv) 12x4 + 8x3 − 6x2 by − 2x2

(i) 5m3 − 30m2 + 45m by 5m

Therefore, the quotient is m2 6m + 9.

(ii) 8x2y2 − 6xy2 + 10x2y3 by 2xy

Therefore, the quotient is 4xy 3y + 5xy2.

(iii) 9x2y − 6xy + 12xy2 by − 3xy

Therefore, the quotient is −3x + 2 4y.

(iv) 12x4 + 8x3 − 6x2 by − 2x2

2-4x+32

Therefore the quotient is −6x2 4x + 3.

#### Question 3:

Write the quotient and remainder when we divide:
(x2 − 4x + 4) by (x − 2)

Therefore, the quotient is $\left(x-2\right)$ and the remainder is 0.

#### Question 4:

Write the quotient and remainder when we divide:
(x2 − 4) by (x + 2)

Therefore, the quotient is $x$−2 and the remainder is 0.

#### Question 5:

Write the quotient and remainder when we divide:
(x2 + 12x + 35) by (x + 7)

(x2 + 12x + 35) by (x + 7)

Therefore, the quotient is $\left(x+5\right)$ and the remainder is 0.

#### Question 6:

Write the quotient and remainder when we divide:
(15x2 + x − 6) by (3x + 2)

Therefore, the quotient is $\left(5x-3\right)$ and the remainder is 0.

#### Question 7:

Write the quotient and remainder when we divide:
(14x2 − 53x + 45) by (7x − 9)

Therefore, the quotient is and the remainder is 0.

#### Question 8:

Write the quotient and remainder when we divide:
(6x2 − 31x + 47) by (2x − 5)

Therefore, the quotient is and the remainder is 7.

#### Question 9:

Write the quotient and remainder when we divide:
(2x3 + x2 − 5x − 2) by (2x + 3)

Therefore, the quotient is $\left({x}^{2}-x-1\right)$ and the remainder is 1.

#### Question 10:

Write the quotient and remainder when we divide:
(x3 + 1) by (x + 1)

Therefore, the quotient is ${x}^{2}$-x+1 and the remainder is 0.

#### Question 11:

Write the quotient and remainder when we divide:
(x4 − 2x3 + 2x2 + x + 4) by (x2 + x + 1)

Therefore, the quotient is ( x2 - 3x + 4) and remainder is 0.

#### Question 12:

Write the quotient and remainder when we divide:
(x3 − 6x2 + 11x − 6) by (x2 − 5x + 6)

Therefore, the quotient is (x-1) and the remainder is 0.

#### Question 13:

Write the quotient and remainder when we divide:
(5x3 − 12x2 + 12x + 13) by (x2 − 3x + 4)

Therefore, the quotient is ( 5x+ 3) and the remainder is (x + 1).

#### Question 14:

Write the quotient and remainder when we divide:
(2x3 − 5x2 + 8x − 5) by (2x2 − 3x + 5)

Therefore, the quotient is (x-1) and the remainder is 0.

#### Question 15:

Write the quotient and remainder when we divide:
(8x4 + 10x3 − 5x2 − 4x + 1) by (2x2 + x − 1)

Therefore, the quotient is ( 4x2+ 3x -2) and the remainder is ( x-1).

#### Question 1:

Find each of the following products:
(i) (x + 6)(x + 6)
(ii) (4x + 5y)(4x + 5y)
(iii) (7a + 9b)(7a + 9b)
(iv) $\left(\frac{2}{3}x+\frac{4}{5}y\right)\left(\frac{2}{3}x+\frac{4}{5}y\right)$
(v) (x2 + 7)(x2 + 7)
(vi) $\left(\frac{5}{6}{a}^{2}+2\right)\left(\frac{5}{6}{a}^{2}+2\right)$

(i) We have:

(ii) We have:

(iii) We have:

(iv) We have:

(v) We have:

(vi) We have:

#### Question 2:

Find each of the following products:
(i) (x − 4)(x − 4)
(ii) (2x − 3y)(2x − 3y)
(iii) $\left(\frac{3}{4}x-\frac{5}{6}y\right)\left(\frac{3}{4}x-\frac{5}{6}y\right)$
(iv) $\left(x-\frac{3}{x}\right)\left(x-\frac{3}{x}\right)$
(v) $\left(\frac{1}{3}{x}^{2}-9\right)\left(\frac{1}{3}{x}^{2}-9\right)$
(vi) $\left(\frac{1}{2}{y}^{2}-\frac{1}{3}y\right)\left(\frac{1}{2}{y}^{2}-\frac{1}{3}y\right)$

(i) We have:

(ii) We have:

(iii) We have:

(iv) We have:

(v) We have:

(vi) We have:

#### Question 3:

Expand:
(i) (8a + 3b)2
(ii) (7x + 2y)2
(iii) (5x + 11)2
(iv) ${\left(\frac{a}{2}+\frac{2}{a}\right)}^{2}$
(v) ${\left(\frac{3x}{4}+\frac{2y}{9}\right)}^{2}$
(vi) (9x − 10)2
(vii) (x2yyz2)2
(viii) ${\left(\frac{x}{y}-\frac{y}{x}\right)}^{2}$
(ix) ${\left(3m-\frac{4}{5}n\right)}^{2}$

We shall use the identities (a+b)2 =a2 +b2 +2ab and (a-b)2 =a2 +b2 -2ab.

(i) We have:
$\left(8a+3b{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(8a\right)}^{2}+2×8a×3b+{\left(3b\right)}^{2}\phantom{\rule{0ex}{0ex}}=64{a}^{2}+48ab+9{b}^{2}$

(ii)We have:
$\left(7x+2y{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(7x\right)}^{2}+2×7x×2y+{\left(2y\right)}^{2}\phantom{\rule{0ex}{0ex}}=49{x}^{2}+28xy+4{y}^{2}$

(iii) We have :
$\left(5x+11{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(5x\right)}^{2}+2×5x×11+{\left(11\right)}^{2}\phantom{\rule{0ex}{0ex}}=25{x}^{2}+110x+121$

(iv) We have:
${\left(\frac{a}{2}+\frac{2}{a}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{a}{2}\right)}^{2}+2×\frac{a}{2}×\frac{2}{a}+{\left(\frac{2}{a}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\frac{a}{4}}^{2}+2+\frac{4}{{a}^{2}}$

(v) We have:
${\left(\frac{3x}{4}+\frac{2y}{9}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{3x}{4}\right)}^{2}+2×\frac{3x}{4}×\frac{2y}{9}+{\left(\frac{2y}{9}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\frac{9x}{16}}^{2}+\frac{1}{3}xy+\frac{4{y}^{2}}{81}$

(vi) We have:
$\left(9x-10{\right)}^{2}\phantom{\rule{0ex}{0ex}}{\left(9x\right)}^{2}-2×9x×10+{\left(10\right)}^{2}\phantom{\rule{0ex}{0ex}}=81{x}^{2}-180x+100$

(vii) We have:
$\left({x}^{2}y-y{z}^{2}{\right)}^{2}\phantom{\rule{0ex}{0ex}}{\left({x}^{2}y\right)}^{2}-2×{x}^{2}y×y{z}^{2}+{\left(y{z}^{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}={x}^{4}{y}^{2}-2{x}^{2}{y}^{2}{z}^{2}+{y}^{2}{z}^{4}$

(viii) We have:
${\left(\frac{x}{y}-\frac{y}{x}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{x}{y}\right)}^{2}-2×\frac{x}{y}×\frac{y}{x}+{\left(\frac{y}{x}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{{y}^{2}}-2+\frac{{y}^{2}}{{x}^{2}}$

(ix) We have:
${\left(3m-\frac{4}{5}n\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(3m\right)}^{2}-2×3m×\frac{4}{5}n+{\left(\frac{4}{5}n\right)}^{2}\phantom{\rule{0ex}{0ex}}=9{m}^{2}-\frac{24mn}{5}+\frac{16}{25}{n}^{2}$

#### Question 4:

Find each of the following products:
(i) (x + 3)(x − 3)
(ii) (2x + 5)(2x − 5)
(iii) (8 + x)(8 − x)
(iv) (7x + 11y)(7x − 11y)
(v) $\left(5{x}^{2}+\frac{3}{4}{y}^{2}\right)\left(5{x}^{2}-\frac{3}{4}{y}^{2}\right)$
(vi) $\left(\frac{4x}{5}-\frac{5y}{3}\right)\left(\frac{4x}{5}+\frac{5y}{3}\right)$
(vii) $\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)$
(viii) $\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}-\frac{1}{y}\right)$

(i) We have:

(ii) We have:

(iii) We have:

(iv) We have:

(v) We have:

(vi) We have:

(vii) We have:

(viii) We have:

(ix) We have:

#### Question 5:

Using the formula for squaring a binomial, evaluate the following:
(i) (54)2
(ii) (82)2
(iii) (103)2
(iv) (704)2

We shall use the identity (a+b)2 =a2 +b2 +2ab.

(i)
${\left(54\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(50+4{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(50\right)}^{2}+2×50×4+{\left(4\right)}^{2}\phantom{\rule{0ex}{0ex}}=2500+400+16\phantom{\rule{0ex}{0ex}}=2916$

(ii)
${\left(82\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(80+2{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(80\right)}^{2}+2×80×2+{\left(2\right)}^{2}\phantom{\rule{0ex}{0ex}}=6400+320+4\phantom{\rule{0ex}{0ex}}=6724$

(iii)
${\left(103\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(100+3{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(100\right)}^{2}+2×100×3+{\left(3\right)}^{2}\phantom{\rule{0ex}{0ex}}=10000+600+9\phantom{\rule{0ex}{0ex}}=10609$

(iv)
${\left(704\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(700+4{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(700\right)}^{2}+2×700×4+{\left(4\right)}^{2}\phantom{\rule{0ex}{0ex}}=490000+5600+16\phantom{\rule{0ex}{0ex}}=495616$

#### Question 6:

Using the formula for squaring a binomial, evaluate the following:
(i) (69)2
(ii) (78)2
(iii) (197)2
(iv) (999)2

We shall use the identity (a-b)2 = a2 +b2 -2ab.

(i)
${\left(69\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(70-1{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(70\right)}^{2}-2×70×1+1\phantom{\rule{0ex}{0ex}}=4900-140+1\phantom{\rule{0ex}{0ex}}=4761\phantom{\rule{0ex}{0ex}}$

(ii)
${\left(78\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(80-2{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(80\right)}^{2}-2×80×2+4\phantom{\rule{0ex}{0ex}}=6400-320+4\phantom{\rule{0ex}{0ex}}=6084\phantom{\rule{0ex}{0ex}}$

(iii)
${\left(197\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(200-3{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(200\right)}^{2}-2×200×3+9\phantom{\rule{0ex}{0ex}}=40000-1200+9\phantom{\rule{0ex}{0ex}}=38809\phantom{\rule{0ex}{0ex}}$

(iv)
${\left(999\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(1000-1{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(1000\right)}^{2}-2×1000×1+1\phantom{\rule{0ex}{0ex}}=1000000-2000+1\phantom{\rule{0ex}{0ex}}=998001\phantom{\rule{0ex}{0ex}}$

#### Question 7:

Find the value of:
(i) (82)2 − (18)2
(ii) (128)2 − (72)2
(iii) 197 × 203
(iv) $\frac{198×198-102×102}{96}$
(v) (14.7 × 15.3)
(vi) (8.63)2 − (1.37)2

We shall use the identity (a-b) (a+b)=a2 - b2.

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

(ii)
$\left(128{\right)}^{2}-\left(72{\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(128-72\right)\left(128+72\right)\phantom{\rule{0ex}{0ex}}=\left(56\right)\left(200\right)\phantom{\rule{0ex}{0ex}}=11200$

(iii)
$197×203\phantom{\rule{0ex}{0ex}}=\left(200-3\right)\left(200+3\right)\phantom{\rule{0ex}{0ex}}={\left(200\right)}^{2}-{\left(3\right)}^{2}\phantom{\rule{0ex}{0ex}}=40000-9\phantom{\rule{0ex}{0ex}}=39991$

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

(v)
$\left(14.7×15.3\right)\phantom{\rule{0ex}{0ex}}=\left(15-0.3\right)×\left(15+0.3\right)\phantom{\rule{0ex}{0ex}}=\left(15{\right)}^{2}-\left(0.3{\right)}^{2}\phantom{\rule{0ex}{0ex}}=225-0.09\phantom{\rule{0ex}{0ex}}=224.91$

(vi)
$\left(8.63{\right)}^{2}-\left(1.37{\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(8.63-1.37\right)\left(8.63+1.37\right)\phantom{\rule{0ex}{0ex}}=\left(7.26\right)\left(10\right)\phantom{\rule{0ex}{0ex}}=72.6$

#### Question 8:

Find the value of the expression (9x2 + 24x + 16), when x = 12.

Therefore, the value of the expression (9x2 + 24x + 16), when x = 12, is 1600.

#### Question 9:

Find the value of the expression (64x2 + 81y2 + 144xy), when x = 11 and $y=\frac{4}{3}.$

Therefore, the value of the expression (64x2 + 81y2 + 144xy), when x = 11 and y = $\frac{4}{3}$, is 10000.y=43

#### Question 10:

Find the value of the expression (36x2 + 25y2 − 60xy), when $x=\frac{2}{3}$ and $y=\frac{1}{5}.$

#### Question 11:

If $\left(x+\frac{1}{x}\right)=4$, find the values of
(i) $\left({x}^{2}+\frac{1}{{x}^{2}}\right)$ and
(ii) $\left({x}^{4}+\frac{1}{{x}^{4}}\right)$.

Therefore, the value of  x2+$\frac{1}{{x}^{2}}$ is 14.

Therefore, the value of x4 + $\frac{1}{{x}^{4}}$ is 194.

#### Question 12:

If $\left(x-\frac{1}{x}\right)=5$, find the values of
(i) $\left({x}^{2}+\frac{1}{{x}^{2}}\right)$
(ii) $\left({x}^{4}+\frac{1}{{x}^{4}}\right).$

#### Question 13:

Find the continued product:
(i) (x + 1)(x − 1)(x2 + 1)
(ii) (x − 3)(x + 3)(x2 + 9)
(iii) (3x − 2y)(3x + 2y)(9x2 + 4y2)
(iv) (2p + 3)(2p − 3)(4p2 + 9)

#### Question 14:

If x + y = 12 and xy = 14, find the value of (x2 + y2).

#### Question 15:

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

#### Question 1:

The sum of (6a + 4bc + 3), (2b − 3c + 4), (11b − 7a + 2c − 1) and (2c − 5a − 6) is
(a) (4a − 6b + 2)
(b) (−3a + 14b − 3c + 2)
(d) (−6a + 17b)
(d) (−6a + 6b + c −4)

(c) (−6a + 17b)

#### Question 2:

(3q + 7p2 − 2r3 + 4) − (4p2 − 2q + 7r3 − 3) = ?
(a) (p2 + 2q + 5r3 + 1)
(b) (11p2 + q + 5r3 + 1)
(c) (−3p2 − 5q + 9r3 − 7)
(d) (3p2 + 5q − 9r3 +7)

(d) (3p2 + 5q − 9r3 +7)

#### Question 3:

(x + 5)(x − 3) = ?
(a) x2 + 5x − 15
(b) x2 − 3x − 15
(c) x2 + 2x + 15
(d) x2 + 2x − 15

(d) x2 + 2x − 15

#### Question 4:

(2x + 3)(3x − 1) = ?
(a) (6x2 + 8x − 3)
(b) (6x2 + 7x − 3)
(c) 6x2 − 7x − 3
(d) (6x2 − 7x + 3)

(b) (6x2 + 7x − 3)

#### Question 5:

(x + 4)(x + 4) = ?
(a) (x2 + 16)
(b) (x2 + 4x + 16)
(c) (x2 + 8x + 16)
(d) (x2 + 16x)

(c) (x2 + 8x + 16)

#### Question 6:

(x − 6)(x − 6) = ?
(a) (x2 − 36)
(b) (x2 + 36)
(c) (x2 − 6x + 36)
(d) (x2 − 12x + 36)

(d) (x2 − 12x + 36)

#### Question 7:

(2x + 5)(2x − 5) = ?
(a) (4x2 + 25)
(b) (4x2 − 25)
(c) (4x2 − 10x + 25)
(d) (4x2 + 10x − 25)

(b) (4x2 − 25)

#### Question 8:

8a2b3 ÷ (−2ab) = ?
(a) 4ab2
(b) 4a2b
(c) −4ab2
(d) −4a2b

(c) −4ab2

#### Question 9:

(2x2 + 3x + 1) ÷ (x + 1) = ?
(a) (x + 1)
(b) (2x + 1)
(c) (x + 3)
(d) (2x + 3)

(b) (2x + 1)

#### Question 10:

(x2 − 4x + 4) ÷ (x − 2) = ?
(a) (x − 2)
(b) (x + 2)
(c) (2 − x)
(d) (2 + x + x2)

(a) (x − 2)

#### Question 11:

(a + 1)(a − 1)(a2 + 1) = ?
(a) (a4 − 2a2 − 1)
(b) (a4a2 − 1)
(c) (a4 − 1)
(d) (a4 + 1)

(c) (a4 − 1)

#### Question 12:

$\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}-\frac{1}{y}\right)=?$
(a) $\left(\frac{1}{{x}^{2}}-\frac{1}{{y}^{2}}\right)$
(b) $\left(\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}}\right)$
(c) $\left(\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}}-\frac{1}{xy}\right)$
(d) $\left(\frac{1}{{x}^{2}}-\frac{1}{{y}^{2}}+\frac{1}{xy}\right)$

a) $\left(\frac{1}{{x}^{2}}-\frac{1}{{y}^{2}}\right)$

(1x21y2)

#### Question 13:

If $\left(x+\frac{1}{x}\right)=5$, then $\left({x}^{2}+\frac{1}{{x}^{2}}\right)=?$
(a) 25
(b) 27
(c) 23
(d) $25\frac{1}{25}$

(c) 23

#### Question 14:

If $\left(x-\frac{1}{x}\right)=6$, then $\left({x}^{2}+\frac{1}{{x}^{2}}\right)=?$
(a) 36
(b) 38
(c) 32
(d) $36\frac{1}{36}$

(b) 38

#### Question 15:

(82)2 − (18)2 = ?
(a) 8218
(b) 6418
(c) 6400
(d) 7204

(c) 6400

[using the identity (a-b)(a+b)=a2 -b2]

#### Question 16:

(197 × 203) = ?
(a) 39991
(b) 39999
(c) 40009
(d) 40001

(a) 39991

$\left(197\right)×\left(203\right)\phantom{\rule{0ex}{0ex}}⇒\left(200-3\right)\left(200+3\right)\phantom{\rule{0ex}{0ex}}⇒{\left(200\right)}^{2}-{\left(3\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒40000-9\phantom{\rule{0ex}{0ex}}⇒39991$                   [using the identity (a+b) (a-b) = a2  -b2]

#### Question 17:

If (a + b) = 12 and ab = 14, then (a2 + b2) = ?
(a) 172
(b) 116
(c) 165
(d) 126

(b) 116

#### Question 18:

If (ab) = 7 and ab = 9, then (a2 + b2) = ?
(a) 67
(b) 31
(c) 40
(d) 58

(a) 67