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#### Question 1:

Which of the following expression are polynomials?
(i) x5 − 2x3 + x + 7
(ii) ${y}^{3}-\sqrt{3}y$
(iii) ${t}^{2}-\frac{2}{5}t+\sqrt{2}$
(iv) $5\sqrt{z}-6$
(v) $x-\frac{1}{x}$
(vi) x108 − 1
(vii) $\sqrt[3]{x}-27$
(viii) $\frac{1}{\sqrt{2}}{x}^{2}-\sqrt{2x}+2$
(ix) ${x}^{-2}+2{x}^{-1}+3$
(x) 1
(xi) $-\frac{3}{5}$
(xii) $\sqrt[3]{2{y}^{2}}-8$
In case of a polynomial, wrtie its degree.

#### Answer:

(i) ${x}^{5}-2{x}^{3}+x+7$ is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 5, so it is a polynomial of degree 5.

(ii) ${y}^{3}-\sqrt{3}y$ is an expression having only non-negative integral powers of y. So, it is a polynomial. Also, the highest power of y is 3, so it is a polynomial of degree 3.

(iii) ${t}^{2}-\frac{2}{5}t+\sqrt{2}$ is an expression having only non-negative integral powers of t. So, it is a polynomial. Also, the highest power of t is 2, so it is a polynomial of degree 2.

(iv) $5\sqrt{z}-6$ is an expression in which $5\sqrt{z}$ has a rational power of z. So, it is not a polynomial.

(v) The given expression may be written as $x-{x}^{-1}$. It contains a term having a negative integral power of x. So, it is not a polynomial.

(vi) ${x}^{108}-1$ is an expression having a non-negative integral power of x. So, it is a polynomial. Also, the power of x is 108, so it is a polynomial of degree 108.

(vii) The given expression may be written as ${x}^{\frac{1}{3}}-27$. It contains a term having a rational power of x. So, it is not a polynomial.

(viii) is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.

(ix) ${x}^{-2}+2{x}^{-1}+3$ is an expression having negative integral powers of x. So, it is not a polynomial.

(x) Clearly, 1 is a constant polynomial of degree 0.

(xi) Clearly, $-\frac{3}{5}$ is a constant polynomial of degree 0.

(xii) is an expression having only a non-negative integral power of x. So, it is a polynomial. Also, the highest power of y is 2, so it is a polynomial of degree 2.

#### Question 2:

Write the degree of each of the following polynomials:
(i) $2x-\sqrt{5}$
(ii) 3 − x + x2− 6x3
(iii) 9
(iv) 8x4 − 36x + 5x7
(v) x9x5  + 3x10 + 8
(vi) 2 − 3x2

#### Answer:

(i) The highest power of x is 1. So, the degree of the given polynomial is 1.

(ii) The highest power of x is 3. So, the degree of the given polynomial is 3.

(iii) Clearly, 9 is a constant polynomial of degree 0.

(iv) The highest power of x is 7. So, the degree of the given polynomial is 7.

(v) The highest power of x is 10. So, the degree of the given polynomial is 10.

(vi) The highest power of x is 2. So, the degree of the given polynomial is 2.

#### Question 3:

Write:
(i) Coefficient of x3 in 2x + x2 − 5x3 + x4
(ii) Coefficient of x in $\sqrt{3}-2\sqrt{2}x+4{x}^{2}$.
(iii) Coefficient of x2 in $\frac{\mathrm{\pi }}{3}{x}^{2}+7x-3$.
(iv) Coefficient of x2 in 3x − 5

#### Answer:

(i) The coefficient of x3 in the given polynomial is $-$5.

(ii) The coefficient of x in the given polynomial is $-2\sqrt{2}$.

(iii) The coefficient of x2 in the given polynomial is $\frac{\mathrm{\pi }}{3}$.

(iv) The coefficient of x2 in the given polynomial is 0.

#### Question 4:

(i) Give an example of a binomial of degree 27.
(ii) Give an example of a monomial of degree 16.
(iii) Give an example of a trinomial of degree 3.

#### Answer:

(i) ${x}^{27}-5$ is a binomial of degree 27.

(ii) x16 is a monomial of degree 16.

(iii) $4{x}^{3}-2x+3$ is a trinomial of degree 3.

#### Question 5:

Classify the following as linear, quadratic and cubic polynomials:
(i) 2x2 + 4x
(ii) x x3
(iii) 2 − yy2
(iv) −7 + z
(v) 5t
(vi) p3

#### Answer:

(i) Clearly, $2{x}^{2}+4x$ is a polynomial of degree 2. So, it is a quadratic polynomial.

(ii) Clearly, $x-{x}^{3}$ is a polynomial of degree 3. So, it is a cubic polynomial.

(iii) Clearly, $2-y-{y}^{2}$ is a polynomial of degree 2. So, it is a quadratic polynomial.

(iv) Clearly, $-$7 + z is a polynomial of degree 1. So, it is a linear polynomial.

(v) Clearly, 5t is a polynomial of degree of 1. So, it is a linear polynomial.

(vi) Clearly, p3 is a polynomial of degree 3. So, it is a cubic polynomial.

#### Question 1:

If p(x) = 5 − 4x + 2x2, find (i) p(0), (ii) p(3), (iii) p(−2)

#### Answer:

$=\left(5-0+0\right)\phantom{\rule{0ex}{0ex}}=5$

$=\left(5-12+18\right)\phantom{\rule{0ex}{0ex}}=11$

$=\left(5+8+8\right)\phantom{\rule{0ex}{0ex}}=21$

#### Question 2:

If p(y) = 4 + 3yy2 + 5y3, find (i) p(0), (ii) p(2), (iii) p(−1).

#### Answer:

$=\left(4+0-0+0\right)\phantom{\rule{0ex}{0ex}}=4$

$=\left(4+6-4+40\right)\phantom{\rule{0ex}{0ex}}=46$

$=\left(4-3-1-5\right)\phantom{\rule{0ex}{0ex}}=-5$

#### Question 3:

If f(t) = 4t2 − 3t + 6, find (i) f(0), (ii) f(4), (iii) f(−5).

#### Answer:

$=\left(0-0+6\right)\phantom{\rule{0ex}{0ex}}=6$

$=\left(64-12+6\right)\phantom{\rule{0ex}{0ex}}=58$

$=\left(100+15+6\right)\phantom{\rule{0ex}{0ex}}=121$

#### Question 4:

Find the zero of the polynomial:
(i) p(x) = x − 5
(ii) q(x) = x + 4
(iii) p(t) = 2t − 3
(iv) f(x) = 3x + 1
(v) g(x) = 5 − 4x
(vi) h(x) = 6x − 1
(vii) p(x) = ax + b, a ≠ 0
(viii) q(x) = 4x
(ix) p(x) = ax, a ≠ 0

#### Question 5:

Verify that:
(i) 4 is a zero of the polynomial p(x) = x − 4.
(ii) −3 is a zero of the polynomial p(x) = x − 3.
(iii) $-\frac{1}{2}$is a zero of the polynomial p(y) = 2y + 1.
(iv) $\frac{2}{5}$is a zero of the polynomial p(x) = 2 − 5x.
(v) 1 and 2 are the zeros of the polynomial p(x) = (x − 1)(x − 2).
(vi) 0 and 3 are the zeros of the polynomial p(x) = x2 − 3x.
(vii) 2 and −3 are the zeros of the polynomial p(x) = x2 + x − 6.

#### Answer:


= 0
Hence, 4 is the zero of the given polynomial.

= 0
Hence, 3 is the zero of the given polynomial.

Hence, $-\frac{1}{2}$ is the zero of the given polynomial.

$=2-2\phantom{\rule{0ex}{0ex}}=0$

Hence, $\frac{2}{5}$ is the zero of the given polynomial.

$=0×\left(-1\right)\phantom{\rule{0ex}{0ex}}=0$
Also,
$p\left(2\right)=\left(2-1\right)\left(2-2\right)$
$=\left(-1\right)×0\phantom{\rule{0ex}{0ex}}=0$

Hence, 1 and 2 are the zeroes of the given polynomial.

Also,

Hence, 0 and 3 are the zeroes of the given polynomial.

$=4-4\phantom{\rule{0ex}{0ex}}=0$
Also,

Hence, 2 and $-3$ are the zeroes of the given polynomial.

#### Question 1:

Using remainder theorem, find the remainder when:
(x3 − 6x2 + 9x + 3) is divided by (x − 1)

#### Answer:

We know:
$x-1=0⇒x=1$
By the remainder theorem, we know that when f(x) is divided by (x $-$ 1), the remainder is f(1).
Thus, we have:

Hence, the required remainder is 7.

#### Question 2:

Using remainder theorem, find the remainder when:
(2x3 − 5x2 + 9x − 8) is divided by (x − 3)

#### Answer:

We have:
$x-3=0⇒x=3$
By the remainder theorem, we know that when f(x) is divided by (x $-$ 3), the remainder is f(3).
Thus, we have:

​Hence, the required remainder is 28.

#### Question 3:

Using remainder theorem, find the remainder when:
(3x4 − 5x2 + 9x − 8) is divided by (x − 3)

#### Answer:

We have:
$x-2=0⇒x=2$
By the remainder theorem, we know that when f(x) is divided by (x $-$ 2), the remainder is f(2).
Thus, we have:

​Hence, the required remainder is 10.

#### Question 4:

Using remainder theorem, find the remainder when:
(x3 − 7x2 + 6x + 4) is divided by (x − 6)

#### Answer:

We have:
$x-6=0⇒x=6$
By the remainder theorem, we know that when f(x) is divided by (x $-$ 6), the remainder is f(6).
Thus, we have:

​Hence, the required remainder is 4.

#### Question 5:

Using remainder theorem, find the remainder when:
(x3 − 6x2 + 13x + 60) is divided by (x + 2)

#### Answer:

We have:
$x+2=0⇒x=-2$
By the remainder theorem, we know that when f(x) is divided by (x + 2), the remainder is f($-$2).
Thus, we have:

Hence, the required remainder is 2.

#### Question 6:

Using remainder theorem, find the remainder when:
(2x4 + 6x3 + 2x2 + x − 8) is divided by (x + 3)

#### Answer:

We have:
$x+3=0⇒x=-3$
By the remainder theorem, we know that when f(x) is divided by (x + 3), the remainder is f($-$3).
Thus, we have:

​Hence, the required remainder is 7.

#### Question 7:

Using remainder theorem, find the remainder when:
(4x3 − 12x2 + 11x − 5) is divided by (2x − 1)

#### Answer:

We have:
$2x-1=0⇒x=\frac{1}{2}$
By the remainder theorem, we know that when f(x) is divided by (2x $-$ 1), the remainder is f($\frac{1}{2}$).
Thus, we have:

Hence, the required remainder is $-$2.

#### Question 8:

Using remainder theorem, find the remainder when:
(81x4 + 54x3 − 9x2 − 3x + 2) is divided by (3x + 2)

#### Answer:

We have:
$3x+2=0⇒x=-\frac{2}{3}$
By the remainder theorem, we know that when f(x) is divided by (3x + 2), the remainder is f($-\frac{2}{3}$).
Thus, we have:

​Hence, the required remainder is 0.

#### Question 9:

Using remainder theorem, find the remainder when:
x3ax2 + 2xa is divided by (xa)

#### Answer:

We have:
$x-a=0⇒x=a$
By the remainder theorem, we know that when f(x) is divided by (x $-$ a), the remainder is f(a).
Thus, we have:

​Hence, the required remainder is a.

#### Question 10:

The polynomials (ax3 + 3x2 − 3) and (2x3 − 5x + a) when divided by (x − 4) leave the same remainder. Find the value of a.

Let:

Now,

We now have:

And,

Thus, we have:

#### Question 11:

The polynomial f(x) = x4 − 2x3 + 3x2ax = b when divided by (x − 1) and (x + 1) leaves the remainder 5 and 19 respectively. Find the values of a and b. Hence, find the remainder when f(x) is divided by (x − 2).

#### Answer:

$\mathrm{Let}:\phantom{\rule{0ex}{0ex}}f\left(x\right)={x}^{4}-2{x}^{3}+3{x}^{2}-ax+b$

Thus, we have:

And,

Now,

$⇒2b=16\phantom{\rule{0ex}{0ex}}⇒b=8$
By putting the value of b, we get the value of a, i.e., 5.
∴ a = 5 and b = 8
Now,

Also,

Thus, we have:

#### Question 1:

Using factor theorem, show that:
(x − 2) is a factor of (x3 − 8)

#### Answer:

Let:
$f\left(x\right)={x}^{3}-8$
Now,
$x-2=0⇒x=2$
By the factor theorem, (x3 $-$ 8) is a factor of the given polynomial if f(2) = 0.
Thus, we have:

Hence, (x $-$ 2) is a factor of the given polynomial.

#### Question 2:

Using factor theorem, show that:
(x − 3) is a factor of (2x3 + 7x2 − 24x − 45)

#### Answer:

Let:
$f\left(x\right)=2{x}^{3}+7{x}^{2}-24x-45$
Now,
$x-3=0⇒x=3$
By the factor theorem, (x $-$ 3) is a factor of the given polynomial if f(3) = 0.
Thus, we have:

Hence, (x $-$ 3) is a factor of the given polynomial.

#### Question 3:

Using factor theorem, show that:
(x − 1) is a factor of (2x4 + 9x3 + 6x2 − 11x − 6)

#### Answer:

Let:
$f\left(x\right)=2{x}^{4}+9{x}^{3}+6{x}^{2}-11x-6$
Here,
$x-1=0⇒x=1$
By the factor theorem, (x $-$ 1) is a factor of the given polynomial if f(1) = 0.
Thus, we have:

Hence, (x $-$ 1) is a factor of the given polynomial.

#### Question 4:

Using factor theorem, show that:
(x + 2) is a factor of (x4x2 − 12)

#### Answer:

Let:
$f\left(x\right)={x}^{4}-{x}^{2}-12$
Here,

By the factor theorem, (x + 2) is a factor of the given polynomial if f ($-$2) = 0.
Thus, we have:

Hence, (x + 2) is a factor of the given polynomial.

#### Question 5:

Using factor theorem, show that:
(x + 5) is a factor of (2x3 + 9x2 − 11x − 30)

#### Answer:

Let:
$f\left(x\right)=2{x}^{3}+9{x}^{2}-11x-30$
Here,
$x+5=0⇒x=-5$
By the factor theorem, (x + 5) is a factor of the given polynomial if f ($-$5) = 0.
Thus, we have:

Hence, (x + 5) is a factor of the given polynomial.

#### Question 6:

Using factor theorem, show that:
(2x − 3) is a factor of (2x4 + x3 − 8x2x + 6)

#### Answer:

Let:
$f\left(x\right)=2{x}^{4}+{x}^{3}-8{x}^{2}-x+6$
Here,
$2x-3=0⇒x=\frac{3}{2}$
By the factor theorem, (2x $-$ 3) is a factor of the given polynomial if $f\left(\frac{3}{2}\right)=0$.
Thus, we have:

Hence, (2x $-$ 3) is a factor of the given polynomial.

#### Question 7:

Using factor theorem, show that:
$\left(x-\sqrt{2}\right)$is a factor of $\left(7{x}^{2}-4\sqrt{2}x-6\right)$

#### Answer:

Let:
$f\left(x\right)=7{x}^{2}-4\sqrt{2}x-6$
Here,
$x-\sqrt{2}=0⇒x=\sqrt{2}$
By the factor theorem, $\left(x-\sqrt{2}\right)$ is a factor of the given polynomial if $f\left(\sqrt{2}\right)=0$
Thus, we have:

Hence, $\left(x-\sqrt{2}\right)$ is a factor of the given polynomial.

#### Question 8:

Using factor theorem, show that:
is a factor of

#### Answer:

Let:
$f\left(x\right)=2\sqrt{2}{x}^{2}+5x+\sqrt{2}$
Here,
$x+\sqrt{2}=0⇒x=-\sqrt{2}$
By the factor theorem, $\left(x+\sqrt{2}\right)$ will be a factor of the given polynomial if $f\left(-\sqrt{2}\right)$ = 0.
Thus, we have:

Hence, $\left(x+\sqrt{2}\right)$ is a factor of the given polynomial.

#### Question 9:

Find the value of k for which (x − 1) is a factor of (2x3 + 9x2 + x + k).

#### Answer:

Let:
$f\left(x\right)=2{x}^{3}+9{x}^{2}+x+k$

Hence, the required value of k is $-$12.

#### Question 10:

Find the value of a for which (x − 4) is a factor of (2x3 − 3x2 − 18x + a).

#### Answer:

Let:
$f\left(x\right)=2{x}^{3}-3{x}^{2}-18x+a$

Hence, the required value of a is $-$8.

#### Question 11:

Find the value of a for which the polynomial (x4x3 − 11x2x + a) is divisible by (x + 3).

#### Answer:

Let:
$f\left(x\right)={x}^{4}-{x}^{3}-11{x}^{2}-x+a$
Now,
$x+3=0⇒x=-3$
By the factor theorem, .
Thus, we have:

Also,

Hence,

#### Question 12:

For what value of a is the polynomial (2x3 + ax2 + 11x + a + 3) exactly divisible by (2x − 1)?

#### Answer:

Let:
$f\left(x\right)=2{x}^{3}+a{x}^{2}+11x+a+3$
Now,
$2x-1=0⇒x=\frac{1}{2}$
By the factor theorem, .
Thus, we have:

Also,

Hence,

#### Question 13:

Find the values of a and b so that the polynomial (x3 − 10x2 + ax + b) is exactly divisible by (x −1) as well as (x − 2).

#### Answer:

Let:
$f\left(x\right)={x}^{3}-10{x}^{2}+ax+b$
Now,
$x-1=0⇒x=1$
By the factor theorem, we can say:
.
Thus, we have:

∴
Also,
$x-2=0⇒x=2$
By the factor theorem, we can say:
.
Thus, we have:

Putting the value of a, we get the value of b, i.e., $-$14.
∴ a = 23 and b = $-$14

#### Question 14:

Find the values of a and b so that the polynomial (x4 + ax3 − 7x2 − 8x + b) is exactly divisible by (x + 2) as well as (x + 3).

#### Answer:

Let:
$f\left(x\right)={x}^{4}+a{x}^{3}-7{x}^{2}-8x+b$
Now,
$x+2=0⇒x=-2$
By the factor theorem, we can say:
.
Thus, we have:

∴
Also,
$x+3=0⇒x=-3$
By the factor theorem, we can say:
.
Thus, we have:

∴

Putting the value of a, we get the value of b, i.e., 12.
∴ a = 2 and b = 12

#### Question 15:

Without actual division, show that (x3 − 3x2 − 13x + 15) is exactly divisible by (x2 + 2x − 3).

#### Answer:

Let:
$f\left(x\right)={x}^{3}-3{x}^{2}-13x+15$
And,
$g\left(x\right)={x}^{2}+2x-3$
$={x}^{2}+x-3x-3\phantom{\rule{0ex}{0ex}}=x\left(x-1\right)+3\left(x-1\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(x+3\right)$
Now, .
For this, we must have:

Thus, we have:

And,

. So, .
Hence, .

#### Question 16:

If (x3 + ax2 + bx + 6) has (x − 2) as a factor and leaves a remainder 3 when divided by (x − 3), find the values of a and b.

#### Answer:

Let:
$f\left(x\right)={x}^{3}+a{x}^{2}+bx+6$

Now,
$x-3=0⇒x=3$
By the factor theorem, we can say:

Now,

Thus, we have:

Subtracting (1) from (2), we get:
a = $-$3
By putting the value of a in (1), we get the value of b, i.e., $-$1.
a = $-$3 and b = $-$1

Factorize:
9x2 + 12xy

#### Answer:

We have:
$9{x}^{2}+12xy\phantom{\rule{0ex}{0ex}}=3x\left(3x+4y\right)$

Factorize:
18x2y − 24xyz

#### Answer:

We have:
$18{x}^{2}y-24xyz\phantom{\rule{0ex}{0ex}}=6xy\left(3y-4z\right)$

Factorize:
27a3b3 − 45a4b2

#### Answer:

We have:
$27{a}^{3}{b}^{3}-45{a}^{4}{b}^{2}\phantom{\rule{0ex}{0ex}}=9{a}^{3}{b}^{2}\left(3b-5a\right)$

#### Question 4:

Factorize:
2a(x + y) − 3b(x + y)

#### Answer:

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

#### Question 5:

Factorize:
2x(p2 + q2) + 4y(p2 + q2)

#### Answer:

We have:
$2x\left({p}^{2}+{q}^{2}\right)+4y\left({p}^{2}+{q}^{2}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\left[x\left({p}^{2}+{q}^{2}\right)+2y\left({p}^{2}+{q}^{2}\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\left({p}^{2}+{q}^{2}\right)\left(x+2y\right)\phantom{\rule{0ex}{0ex}}$

#### Question 6:

Factorize:
x(a − 5) + y(5 − a)

#### Answer:

We have:
$x\left(a-5\right)+y\left(5-a\right)=x\left(a-5\right)-y\left(a-5\right)$
$=\left(a-5\right)\left(x-y\right)$

#### Question 7:

Factorize:
4(a + b) − 6(a + b)2

#### Answer:

We have:
$4\left(a+b\right)-6{\left(a+b\right)}^{2}=2\left(a+b\right)\left[2-3\left(a+b\right)\right]$
$=2\left(a+b\right)\left(2-3a-3b\right)$

#### Question 8:

Factorize:
8(3a − 2b)2 − 10(3a − 2b)

#### Answer:

We have:
$8{\left(3a-2b\right)}^{2}-10\left(3a-2b\right)=2\left(3a-2b\right)\left[4\left(3a-2b\right)-5\right]$
$=2\left(3a-2b\right)\left(12a-8b-5\right)$

#### Question 9:

Factorize:
x(x + y)3 − 3x2y(x + y)

#### Answer:

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

#### Question 10:

Factorize:
x3 + 2x2 + 5x + 10

#### Answer:

We have:
${x}^{3}+2{x}^{2}+5x+10=\left({x}^{3}+2{x}^{2}\right)+\left(5x+10\right)$
$={x}^{2}\left(x+2\right)+5\left(x+2\right)\phantom{\rule{0ex}{0ex}}=\left(x+2\right)\left({x}^{2}+5\right)$

#### Question 11:

Factorize:
x2 + xy − 2xz − 2yz

We have:

#### Question 12:

Factorize:
a3ba2b + 5ab − 5b

#### Answer:

We have:

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

#### Question 13:

Factorize:
8 − 4a − 2a3 + a4

#### Answer:

We have:

#### Question 14:

Factorize:
x3 − 2x2y + 3xy2 − 6y3

#### Answer:

We have:
${x}^{3}-2{x}^{2}y+3x{y}^{2}-6{y}^{3}=\left({x}^{3}-2{x}^{2}y\right)+\left(3x{y}^{2}-6{y}^{3}\right)$
$={x}^{2}\left(x-2y\right)+3{y}^{2}\left(x-2y\right)\phantom{\rule{0ex}{0ex}}=\left(x-2y\right)\left({x}^{2}+3{y}^{2}\right)$

#### Question 15:

Factorize:
px − 5q + pq − 5x

#### Answer:

We have:
$px-5q+pq-5x=\left(px-5x\right)+\left(pq-5q\right)$
$=x\left(p-5\right)+q\left(p-5\right)\phantom{\rule{0ex}{0ex}}=\left(p-5\right)\left(x+q\right)$

Factorize:
x2 + yxyx

#### Answer:

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

#### Question 17:

Factorize:
(3a − 1)2 − 6a + 2

#### Answer:

We have:
${\left(3a-1\right)}^{2}-6a+2={\left(3a-1\right)}^{2}-2\left(3a-1\right)$
$=\left(3a-1\right)\left[\left(3a-1\right)-2\right]\phantom{\rule{0ex}{0ex}}=\left(3a-1\right)\left(3a-1-2\right)\phantom{\rule{0ex}{0ex}}=\left(3a-1\right)\left(3a-3\right)\phantom{\rule{0ex}{0ex}}=3\left(3a-1\right)\left(a-1\right)$

#### Question 18:

Factorize:
(2x − 3)2 − 8x + 12

#### Answer:

We have:
${\left(2x-3\right)}^{2}-8x+12={\left(2x-3\right)}^{2}-4\left(2x-3\right)$
$=\left(2x-3\right)\left[\left(2x-3\right)-4\right]\phantom{\rule{0ex}{0ex}}=\left(2x-3\right)\left(2x-3-4\right)\phantom{\rule{0ex}{0ex}}=\left(2x-3\right)\left(2x-7\right)$

Factorize:
a3 + a − 3a2 − 3

#### Answer:

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

#### Question 20:

Factorize:
3ax − 6ay − 8by + 4bx

#### Answer:

We have:
$3ax-6ay-8by+4bx=\left(3ax-6ay\right)+\left(4bx-8by\right)$
$=3a\left(x-2y\right)+4b\left(x-2y\right)\phantom{\rule{0ex}{0ex}}=\left(x-2y\right)\left(3a+4b\right)$

#### Question 21:

Factorize:
abx2 + a2x + b2x + ab

#### Answer:

We have:
$ab{x}^{2}+{a}^{2}x+{b}^{2}x+ab=\left(ab{x}^{2}+{b}^{2}x\right)+\left({a}^{2}x+ab\right)$
$=bx\left(ax+b\right)+a\left(ax+b\right)\phantom{\rule{0ex}{0ex}}=\left(ax+b\right)\left(bx+a\right)$

#### Question 22:

Factorize:
x3x2 + ax + xa − 1

#### Answer:

We have:
${x}^{3}-{x}^{2}+ax+x-a-1=\left({x}^{3}-{x}^{2}\right)+\left(ax-a\right)+\left(x-1\right)$
$={x}^{2}\left(x-1\right)+a\left(x-1\right)+1\left(x-1\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left({x}^{2}+a+1\right)$

#### Question 23:

Factorize:
2x + 4y − 8xy − 1

#### Answer:

We have:
$2x+4y-8xy-1=\left(2x-8xy\right)-\left(1-4y\right)$
$=2x\left(1-4y\right)-1\left(1-4y\right)\phantom{\rule{0ex}{0ex}}=\left(1-4y\right)\left(2x-1\right)$

#### Question 24:

Factorize:
ab(x2 + y2) − xy(a2 + b2)

#### Answer:

We have:
$ab\left({x}^{2}+{y}^{2}\right)-xy\left({a}^{2}+{b}^{2}\right)=ab{x}^{2}+ab{y}^{2}-{a}^{2}xy-{b}^{2}xy$
$=\left(ab{x}^{2}-{a}^{2}xy\right)-\left({b}^{2}xy-ab{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=ax\left(bx-ay\right)-by\left(bx-ay\right)\phantom{\rule{0ex}{0ex}}=\left(bx-ay\right)\left(ax-by\right)$

#### Question 25:

Factorize:
a2 + ab(b + 1) + b3

#### Answer:

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

#### Question 26:

Factorize:
a3 + ab(1 − 2a) − 2b2

#### Answer:

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

#### Question 27:

Factorize:
2a2 + bc − 2abac2

#### Answer:

We have:
$2{a}^{2}+bc-2ab-ac=\left(2{a}^{2}-2ab\right)-\left(ac-bc\right)$
$=2a\left(a-b\right)-c\left(a-b\right)\phantom{\rule{0ex}{0ex}}=\left(a-b\right)\left(2a-c\right)$

#### Question 28:

Factorize:
(ax + by)2 + (bxay)2

#### Answer:

We have:
${\left(ax+by\right)}^{2}+{\left(bx-ay\right)}^{2}=\left[{\left(ax\right)}^{2}+2×ax×by+{\left(by\right)}^{2}\right]+\left[{\left(bx\right)}^{2}-2×bx×ay+{\left(ay\right)}^{2}\right]$
$={a}^{2}{x}^{2}+2abxy+{b}^{2}{y}^{2}+{b}^{2}{x}^{2}-2abxy+{a}^{2}{y}^{2}\phantom{\rule{0ex}{0ex}}={a}^{2}{x}^{2}+{b}^{2}{y}^{2}+{b}^{2}{x}^{2}+{a}^{2}{y}^{2}\phantom{\rule{0ex}{0ex}}=\left({a}^{2}{x}^{2}+{b}^{2}{x}^{2}\right)+\left({a}^{2}{y}^{2}+{b}^{2}{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left({a}^{2}+{b}^{2}\right)+{y}^{2}\left({a}^{2}+{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left({a}^{2}+{b}^{2}\right)\left({x}^{2}+{y}^{2}\right)$

Factorize:
a(a + bc) − bc

#### Answer:

We have:
$a\left(a+b-c\right)-bc={a}^{2}+ab-ac-bc$
$=\left({a}^{2}-ac\right)+\left(ab-bc\right)\phantom{\rule{0ex}{0ex}}=a\left(a-c\right)+b\left(a-c\right)\phantom{\rule{0ex}{0ex}}=\left(a-c\right)\left(a+b\right)$

Factorize:
a(a − 2bc) + 2bc

#### Answer:

We have:
$a\left(a-2b-c\right)+2bc={a}^{2}-2ab-ac+2bc$
$=\left({a}^{2}-2ab\right)-\left(ac-2bc\right)\phantom{\rule{0ex}{0ex}}=a\left(a-2b\right)-c\left(a-2b\right)\phantom{\rule{0ex}{0ex}}=\left(a-2b\right)\left(a-c\right)$

#### Question 31:

Factorize:
a2x2 + (ax2 + 1)x + a

#### Answer:

We have:
${a}^{2}{x}^{2}+\left(a{x}^{2}+1\right)x+a=\left(a{x}^{2}+1\right)x+\left({a}^{2}{x}^{2}+a\right)$
$=x\left(a{x}^{2}+1\right)+a\left(a{x}^{2}+1\right)\phantom{\rule{0ex}{0ex}}=\left(a{x}^{2}+1\right)\left(x+a\right)$

#### Question 32:

Factorize:
ab(x2 + 1) + x(a2 + b2)

#### Answer:

We have:
$ab\left({x}^{2}+1\right)+x\left({a}^{2}+{b}^{2}\right)=ab{x}^{2}+ab+{a}^{2}x+{b}^{2}x$
$=\left(ab{x}^{2}+{a}^{2}x\right)+\left({b}^{2}x+ab\right)\phantom{\rule{0ex}{0ex}}=ax\left(bx+a\right)+b\left(bx+a\right)\phantom{\rule{0ex}{0ex}}=\left(bx+a\right)\left(ax+b\right)$

#### Question 33:

Factorize:
x2 − (a + b)x + ab

#### Answer:

We have:
${x}^{2}-\left(a+b\right)x+ab={x}^{2}-ax-bx+ab$
$=\left({x}^{2}-ax\right)-\left(bx-ab\right)\phantom{\rule{0ex}{0ex}}=x\left(x-a\right)-b\left(x-a\right)\phantom{\rule{0ex}{0ex}}=\left(x-a\right)\left(x-b\right)$

#### Question 34:

Factorize:
${x}^{2}+\frac{1}{{x}^{2}}-2-3x+\frac{3}{x}$

Factorize:
25x2 − 64y2

Factorize:
100 − 9x2

Factorize:
5x2 − 7y2

Factorize:
(3x + 5y2) − 4z2

Factorize:
150 − 6x2

Factorize:
20x2 − 45

Factorize:
3x3 − 48x

Factorize:
2 − 50x2

Factorize:
27a2 − 48b2

Factorize:
x − 64x3

Factorize:
8ab2 − 18a3

Factorize:
3a3b − 243ab3

Factorize:
(a + b)3ab

Factorize:
108a2 − 3(bc)2

Factorize:
x3 − 5x2x + 5

#### Question 16:

Factorize:
a2 + 2ab + b2 − 9c2

Factorize:
9 − a2 + 2abb2

Factorize:
a2b2 − 4ac + 4c2

#### Question 19:

Factorize:
9a2 + 3a − 8b − 64b2

Factorize:
x2y2 + 6y − 9

#### Question 21:

Factorize:
4x2 − 9y2 − 2x − 3y

Factorize:
x4 − 1

Factorize:
aba2 + b2

Factorize:
x4 − 625

Factorize:
x2 + 11x + 30

#### Answer:

We have:
${x}^{2}+11x+30$
We have to split 11 into two numbers such that their sum of is 11 and their product is 30.
Clearly, .

Factorize:
x2 + 18x + 32

#### Answer:

We have:
${x}^{2}+18x+32$
We have to split 18 into two numbers such that their sum is 18 and their product is 32.
Clearly, .

Factorize:
x2 + 7x − 18

#### Answer:

We have:
${x}^{2}+7x-18$
We have to split 7 into two numbers such that their sum is 7 and their product is ($-$18).
Clearly, .

Factorize:
x2 + 5x − 6

#### Answer:

We have:
${x}^{2}+5x-6$
We have to split 5 into two numbers such that their sum is 5 and their product is ($-$6).
Clearly, .

Factorize:
y2 − 4y + 3

#### Answer:

We have:
${y}^{2}-4y+3$
We have to split ($-$4) into two numbers such that their sum is ($-$4) and their product is 3.
Clearly, .

Factorize:
x2 − 21x + 108

#### Answer:

We have:
${x}^{2}-21x+108$
We have to split ($-$21) into two numbers such that their sum is ($-$21) and their product is 108.
Clearly, .

Factorize:
x2 − 11x − 80

#### Answer:

We have:
${x}^{2}-11x-80$
We have to split ($-$11) into two numbers such that their sum is ($-$11) and their product is ($-$80).
Clearly, .

Factorize:
x2x − 156

#### Answer:

We have:
${x}^{2}-x-156$
We have to split ($-$1) into two numbers such that their sum is ($-$1) and their product is ($-$156).
Clearly, .

Factorize:
z2 − 32z − 105

#### Answer:

We have:
${z}^{2}-32z-105$
We have to split ($-$32) into two numbers such that their sum is ($-$32) and their product is ($-$105).
Clearly, .

Factorize:
40 + 3xx2

#### Answer:

We have to split ($-$3) into two numbers such that their sum is ($-$3) and their product is ($-$40).
Clearly, .

Factorize:
6 − xx2

#### Answer:

We have to split 1 into two numbers such that their sum is 1 and their product is ($-$6).
Clearly, .

Factorize:
7x2 + 49x + 84

#### Answer:

We now have to split 7 into two numbers such that their sum is 7 and their product is 12.
Clearly, .

Factorize:
m2 + 17mn − 84n2

#### Answer:

We have:
${m}^{2}+17mn-84{n}^{2}$
We have to split 17mn into two terms such that their sum is 17mn and their product is ($-$84 m2n2).
Clearly, .

Factorize:
5x2 + 16x + 3

#### Answer:

We have:
$5{x}^{2}+16x+3$
We have to split 16 into two numbers such that their sum is 16 and their product is 15, i.e., $5×3$.
Clearly, .

Factorize:
6x2 + 17x + 12

#### Answer:

We have:
$6{x}^{2}+17x+12$
We have to split 17 into two numbers such that their sum is 17 and their product is 72, i.e., $6×12$.
Clearly, .

Factorize:
9x2 + 18x + 8

#### Answer:

We have:
$9{x}^{2}+18x+8$
We have to split 18 into two numbers such that their sum is 18 and their product is 72, i.e., $9×8$.
Clearly, .

Factorize:
14x2 + 9x + 1

#### Answer:

We have:
$14{x}^{2}+9x+1$

We have to split 9 into two numbers such that their sum is 9 and their product is 14, i.e., $14×1$.
Clearly, .

Factorize:
2x2 + 3x − 90

#### Answer:

We have:
$2{x}^{2}+3x-90$
We have to split 3 into two numbers such that their sum is 3 and their product is ($-$180), i.e., $2×\left(-90\right)$.
Clearly, .

Factorize:
2x2 + 11x − 21

#### Answer:

We have:
$2{x}^{2}+11x-21$
We have to split 11 into two numbers such that their sum is 11 and their product is ($-$42), i.e., $2×\left(-21\right)$.
Clearly, .

Factorize:
3x2 − 14x + 8

#### Answer:

We have:
$3{x}^{2}-14x+8$
We have to split ($-$14) into two numbers such that their sum is ($-$14) and their product is 24, i.e., $3×8$.
Clearly, .

Factorize:
18x2 + 3x − 10

#### Answer:

We have:
$18{x}^{2}+3x-10$
We have to split 3 into two numbers such that their sum is 3 and their product is ($-$180), i.e., $18×\left(-10\right)$.
Clearly, .

Factorize:
15x2 + 2x − 8

#### Answer:

We have:
$15{x}^{2}+2x-8$
We have to split 2 into two numbers such that their sum is 2 and their product is ($-$120), i.e., $15×\left(-8\right)$.
Clearly, .

Factorize:
6x2 + 11x − 10

#### Answer:

We have:
$6{x}^{2}+11x-10$
We have to split 11 into two numbers such that their sum is 11 and their product is ($-$60), i.e., $6×\left(-10\right)$.
Clearly, .

Factorize:
30x2 + 7x − 15

#### Answer:

We have:
$30{x}^{2}+7x-15$

We have to split 7 into two numbers such that their sum is 7 and their product is ($-$450), i.e., $30×\left(-15\right)$.
Clearly, .

Factorize:
24x2 − 41x + 12

#### Answer:

We have:
$24{x}^{2}-41x+12$
We have to split ($-$41) into two numbers such that their sum is ($-$41) and their product is 288, i.e., $24×12$.
Clearly, .

Factorize:
2x2 − 7x − 15

#### Answer:

We have:
$2{x}^{2}-7x-15$
We have to split ($-$7) into two numbers such that their sum is ($-$7) and their product is ($-$30), i.e., $2×\left(-15\right)$.
Clearly, .

Factorize:
6x2 − 5x − 21

#### Answer:

We have:
$6{x}^{2}-5x-21$
We have to split ($-$5) into two numbers such that their sum is ($-$5) and their product is ($-$126), i.e., $6×\left(-21\right)$.
Clearly, .

Factorize:
10x2 − 9x − 7

#### Answer:

We have:
$10{x}^{2}-9x-7$

We have to split ($-$9) into two numbers such that their sum is ($-$9) and their product is ($-$70), i.e., $10×\left(-7\right)$.
Clearly, .

Factorize:
5x2 − 16x − 21

#### Answer:

We have:
$5{x}^{2}-16x-21$
We have to split ($-$16) into two numbers such that their sum is ($-$16) and their product is ($-$105), i.e., $5×\left(-21\right)$.
Clearly, .

Factorize:
2x2x − 21

#### Answer:

We have:
$2{x}^{2}-x-21$
We have to split ($-$1) into two numbers such that their sum is ($-$1) and their product is ($-$42), i.e., $2×\left(-21\right)$.
Clearly, .

Factorize:
15x2x − 128

#### Answer:

We have:
$15{x}^{2}-x-28$
We have to split ($-$1) into two numbers such that their sum is ($-$1) and their product is ($-$420), i.e., $15×\left(-28\right)$.
Clearly, .

Factorize:
8a2 − 27ab + 9b2

#### Answer:

We have:
$8{a}^{2}-27ab+9{b}^{2}$

We have to split ($-$27) into two numbers such that their sum is ($-$27) and their product is 72, i.e., $8×9$.
Clearly, .

#### Question 33:

Factorize:
5x2 + 33xy − 14y2

#### Answer:

We have:
$5{x}^{2}+33xy-14{y}^{2}$
We have to split 33 into two numbers such that their sum is 33 and their product is ($-$70), i.e., $5×\left(-14\right)$.
Clearly, .

Factorize:
3x3x2 − 10x

#### Answer:

We have to split ($-$1) into two numbers such that their sum is ($-$1) and their product is ($-$30), i.e., $3×\left(-10\right)$.
Clearly, .

#### Question 35:

Factorize:
$\frac{1}{3}{x}^{2}-2x-9$

#### Answer:

Now, we have to split ($-$6) into two numbers such that their sum is ($-$6) and their product is ($-$27).
Clearly, .

#### Question 36:

Factorize:
${x}^{2}-2x+\frac{7}{16}$

#### Answer:

Now, we have to split ($-$32) into two numbers such that their sum is ($-$32) and their product is 112, i.e., $16×7$.
Clearly, .

#### Question 37:

Factorize:
$\sqrt{2}{x}^{2}+3x+\sqrt{2}$

#### Answer:

We have:
$\sqrt{2}{x}^{2}+3x+\sqrt{2}$
We have to split 3 into two numbers such that their sum is 3 and their product is 2, i.e., $\sqrt{2}×\sqrt{2}$.
Clearly, .

#### Question 38:

Factorize:
$\sqrt{5}{x}^{2}+2x-3\sqrt{5}$

#### Answer:

We have:
$\sqrt{5}{x}^{2}+2x-3\sqrt{5}$
We have to split 2 into two numbers such that their sum is 2 and product is ($-$15), i.e.,$\sqrt{5}×\left(-3\sqrt{5}\right)$.
Clearly, .

#### Question 39:

Factorize:
$2{x}^{2}+3\sqrt{3}x+3$

#### Answer:

We have:
$2{x}^{2}+3\sqrt{3}x+3$
We have to split $3\sqrt{3}$ into two numbers such that their sum is $3\sqrt{3}$ and their product is 6, i.e.,$2×3$.
Clearly, .

#### Question 40:

Factorize:
$2\sqrt{3}{x}^{2}+x-5\sqrt{3}$

#### Answer:

We have:
$2\sqrt{3}{x}^{2}+x-5\sqrt{3}$
We have to split 1 into two numbers such that their sum is 1 and product is 30, i.e.,$2\sqrt{3}×\left(-5\sqrt{3}\right)$.
Clearly, .

#### Question 41:

Factorize:
$5\sqrt{5}{x}^{2}+20x+3\sqrt{5}$

#### Answer:

We have:
$5\sqrt{5}{x}^{2}+20x+3\sqrt{5}$
We have to split 20 into two numbers such that their sum is 20 and their product is 75.
Clearly,

#### Question 42:

Factorize:
$7\sqrt{2}{x}^{3}-10x-4\sqrt{2}$

#### Answer:

We have:
$7\sqrt{2}{x}^{2}-10x-4\sqrt{2}$
We have to split ($-$10) into two numbers such that their sum is ($-$10) and their product is ($-$56), i.e., $7\sqrt{2}×\left(-4\sqrt{2}\right)=-28×2$.
Clearly, .

#### Question 43:

Factorize:
$6\sqrt{3}{x}^{2}-47x+5\sqrt{3}$

#### Answer:

We have:
$6\sqrt{3}{x}^{2}-47x+5\sqrt{3}$
Now, we have to split ($-$47) into two numbers such that their sum is ($-$47) and their product is 90.
Clearly, .

#### Question 44:

Factorize:
$7{x}^{2}+2\sqrt{14}x+2$

#### Answer:

We have:
$7{x}^{2}+2\sqrt{14}x+2$
We have to split $2\sqrt{14}$ into two numbers such that their sum is $2\sqrt{14}$ and product is 14.
Clearly, .

#### Question 45:

Factorize:
2(x + y)2 − 9(x + y) − 5

#### Answer:

We have:
$2{\left(x+y\right)}^{2}-9\left(x+y\right)-5\phantom{\rule{0ex}{0ex}}\mathrm{Let}:\phantom{\rule{0ex}{0ex}}\left(x+y\right)=u\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Thus, the given expression becomes
$2{u}^{2}-9u-5\phantom{\rule{0ex}{0ex}}$
Now, we have to split ($-$9) into two numbers such that their sum is ($-$9) and their product is ($-$10).
Clearly, .

Putting $u=\left(x+y\right)$, we get:

#### Question 46:

Factorize:
9(2ab)2 − 4(2ab) − 13

#### Answer:

We have:
$9\left(2a-b{\right)}^{2}-4\left(2a-b\right)-13\phantom{\rule{0ex}{0ex}}\mathrm{Let}:\phantom{\rule{0ex}{0ex}}\left(2a-b\right)=p\phantom{\rule{0ex}{0ex}}$
Thus, the given expression becomes
$9{p}^{2}-4p-13\phantom{\rule{0ex}{0ex}}$
Now, we must split ($-$4) into two numbers such that their sum is ($-$4) and their product is ($-$117).
Clearly, .

Putting $p=\left(2a-b\right)$, we get:

#### Question 47:

Factorize:
7(x − 2y)2 − 25(x − 2y) + 12

#### Answer:

We have:
$7\left(x-2y{\right)}^{2}-25\left(x-2y\right)+12\phantom{\rule{0ex}{0ex}}\mathrm{Let}:\phantom{\rule{0ex}{0ex}}\left(x-2y\right)=a\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$\therefore 7{\left(x-2y\right)}^{2}-25\left(x-2y\right)+12=7{a}^{2}-25a+12\phantom{\rule{0ex}{0ex}}$
Now, we have to split ($-$25) into two numbers such that their sum is ($-$25) and their product is 84.
Clearly, .

Putting $a=x-2y$, we get:

Factorize:
4x4 + 7x2 − 2

#### Answer:

Substituting $a={x}^{2}$, we get:

#### Question 1:

Expand:
(i) (a + 2b + 5c)2
(ii) (2bb + c)2
(iii) (a − 2b − 3c)2

#### Question 2:

Expand:
(i) (2a − 5b − 7c)2
(ii) (−3a + 4b − 5c)2
(iii) ${\left(\frac{1}{2}a-\frac{1}{4}a+2\right)}^{2}$

#### Question 3:

Factorize: 4x2 + 9y2 + 16z2 + 12xy − 24yz − 16xz.

#### Question 4:

Factorize: 9x2 + 16y2 + 4z2 − 24xy + 16yz − 12xz

#### Question 5:

Factorize: 25x2 + 4y2 + 9z2 − 20xy − 12yz + 30xz.

Evaluate:
(i) (99)2
(ii) (998)2

#### Question 1:

Expand:
(i) (3x + 2)3
(ii) (3a − 2b)3
(iii) ${\left(\frac{2}{3}x+1\right)}^{3}$

#### Question 2:

Expand:
(i) ${\left(2x-\frac{2}{x}\right)}^{3}$
(ii) ${\left(3a+\frac{1}{4b}\right)}^{3}$
(iii) ${\left(\frac{4}{5}x-2\right)}^{3}$

Evaluate:
(i) (95)3
(ii) (999)3

Factorize:
x3 + 27

Factorize:
8x3 + 27y3

Factorize:
343 + 125b3

Factorize:
1 + 64x3

#### Question 5:

Factorize:
$125{a}^{3}+\frac{1}{8}$

#### Question 6:

Factorize:
$216{x}^{3}+\frac{1}{125}$

Factorize:
16x4 + 54x

Factorize:
7a3 + 56b3

Factorize:
x5 + x2

Factorize:
a3 + 0.008

Factorize:
x6 + y6

#### Question 12:

Factorize:
2a3 + 16b3 − 5a − 10b

Factorize:
x3 − 512

Factorize:
64x3 − 343

Factorize:
1 − 27x3

Factorize:
x3 − 125y3

#### Question 17:

Factorize:
$8{x}^{3}-\frac{1}{27{y}^{3}}$

Factorize:
a3 − 0.064

Factorize:
(a + b)3 − 8

Factorize:
x6 − 729

Factorize:
(a + b)3 − (ab)3

Factorize:
x − 8xy3

Factorize:
32x4 − 500x

Factorize:
3a7b − 81a4b4

#### Question 25:

Factorize:
${a}^{3}-\frac{1}{{a}^{3}}-2a+\frac{2}{a}$

#### Question 26:

Factorize:
8a3b3 − 4ax + 2bx

#### Question 27:

Factorize:
a3 + 3a2b + 3ab2 + b3 − 8

#### Question 1:

Factorize:
125a3 + b3 + 64c3 − 60abc

#### Answer:

#### Question 2:

Factorize:
a3 + 8b3 + 64c3 − 24abc

#### Question 3:

Factorize:
1 + b3 + 8c3 − 6bc

#### Question 4:

Factorize:
216 + 27b3 + 8c3 − 108abc

#### Question 5:

Factorize:
27a3b3 + 8c3 + 18abc

#### Question 6:

Factorize:
8a3 + 125b3 − 64c3 + 120abc

#### Question 7:

Factorize:
8 − 27b3 − 343c3 − 126bc

#### Question 8:

Factorize:
125 − 8x3 − 27y3 − 90xy

Factorize:

#### Question 10:

Factorize:
x3 + y3 − 12xy + 64

#### Question 11:

Factorize:
(ab)3 + (bc)3 + (ca)3

#### Answer:

${\left(a-b\right)}^{3}+{\left(b-c\right)}^{3}+{\left(c-a\right)}^{3}$

#### Question 12:

Factorize:
(3a − 2b)3 + (2b − 5c)3 + (5c − 3a)3

#### Question 13:

Factorize:
a3(bc)3 + b3(ca)3 + c3(ab)3

#### Question 14:

Factorize:
(5a − 7b)3 + (9c − 5a)3 + (7b − 9c)3

#### Question 15:

Find the product:
(x + yz) (x2 + y2 + z2xy + yz + zx)

#### Question 16:

Find the product:
(x − 2y + 3) (x2 + 4y2 + 2xy − 3x + 6y + 9)

#### Question 17:

Find the product:
(x − 2y z) (x2 + 4y2 + z2 + 2xy + zx − 2yz)

#### Question 18:

If x + y + 4 = 0, find the value of (x3 + y3 − 12xy + 64).

#### Question 19:

If x = 2y + 6, find the value of (x3 − 8y3 − 36xy − 216)

#### Question 1:

Which of the following expressions is a polynomial in one variable?
(a) $x+\frac{2}{x}+3$
(b) $3\sqrt{x}+\frac{2}{\sqrt{x}}+5$
(c)
(d) x10 + y5 + 8

#### Answer:

(c)

Clearly, $\sqrt{2}{x}^{2}-\sqrt{3}x+6$ is a polynomial in one variable because it has only non-negative integral powers of x.

#### Question 2:

Which of the following expression is a polynomial?
(a) $\sqrt{x}-1$
(b) $\frac{x-1}{x+1}$
(c) ${x}^{2}-\frac{2}{{x}^{2}}+5$
(d) ${x}^{2}+\frac{2{x}^{3/2}}{\sqrt{x}}+6$

#### Answer:

(d) ${x}^{2}+\frac{2{x}^{3/2}}{\sqrt{x}}+6$

We have:

${x}^{2}+\frac{2{x}^{\frac{3}{2}}}{\sqrt{x}}+6={x}^{2}+2{x}^{\frac{3}{2}}{x}^{-\frac{1}{2}}+6$
$={x}^{2}+2x+6$

It a polynomial because it has only non-negative integral powers of x.

#### Question 3:

(a) $\sqrt[3]{y}+4$
(b) $\sqrt{y}-3$
(c) y
(d) $\frac{1}{\sqrt{y}}+7$

#### Answer:

(c) y

y is a polynomial because it has a non-negative integral power 1.

#### Question 4:

Which of the following is a polynimial?
(a) $x-\frac{1}{x}+2$
(b) $\frac{1}{x}+5$
(c) $\sqrt{x}+3$
(d) −4

#### Answer:

(d) −4

$-$4 is a constant polynomial of degree zero.

#### Question 5:

Which of the following is a polynomial?
(a) x−2 + x−1 + 3
(b) x + x−1 + 2
(c) x−1
(d) 0

#### Answer:

(d) 0

0 is a polynomial whose degree is not defined.

#### Question 6:

Which of the following is quadratic polynomial?
(a) x + 4
(b) x3 + x
(c) x3 + 2x + 6
(d) x2 + 5x + 4

#### Answer:

(d) x2 + 5x + 4

${x}^{2}+5x+4$ is a polynomial of degree 2. So, it is a quadratic polynomial.

#### Question 7:

Which of the following is a linear polynomial?
(a) x + x2
(b) x + 1
(c) 5x2x + 3
(d) $x+\frac{1}{x}$

#### Answer:

(b) x + 1

Clearly, $x+1$ is a polynomial of degree 1. So, it is a linear polynomial.

#### Question 8:

Which of the following is a binomial?
(a) x2 + x + 3
(b) x2 + 4
(c) 2x2
(d) $x+3+\frac{1}{x}$

#### Answer:

(b) x2 + 4

Clearly, ${x}^{2}+4$ is an expression having two non-zero terms. So, it is a binomial.

#### Question 9:

$\sqrt{3}$is a polynomial of degree
(a) $\frac{1}{2}$
(b) 2
(c) 1
(d) 0

#### Answer:

(d) 0

$\sqrt{3}$ is a constant term, so it is a polynomial of degree 0.

#### Question 10:

Degree of the zero polynomial is
(a) 1
(b) 0
(c) not defined
(d) none of these

#### Answer:

(c) not defined

Degree of the zero polynomial is not defined.

#### Question 11:

Zero of the polynomial p(x) = 2x + 3 is
(a) $\frac{3}{2}$
(b) $\frac{-3}{2}$
(c) $\frac{-2}{3}$
(d) $\frac{1}{2}$

#### Answer:

(b) $\frac{-3}{2}$

Zero of the polynomial p(x) is given by p(x) = 0.
Thus, we have:
$2x+3=0\phantom{\rule{0ex}{0ex}}⇒2x=-3\phantom{\rule{0ex}{0ex}}⇒x=\frac{-3}{2}$

#### Question 12:

Zero is the polynomial p(x) = 2 − 5x is
(a) $\frac{2}{5}$
(b) $\frac{5}{2}$
(c) $\frac{-2}{5}$
(d) $\frac{-5}{2}$

#### Answer:

(a) $\frac{2}{5}$

Zero of the polynomial p(x) is given by p(x) = 0.
Thus, we have:
$2-5x=0\phantom{\rule{0ex}{0ex}}⇒2=5x\phantom{\rule{0ex}{0ex}}⇒x=\frac{2}{5}$

#### Question 13:

Zero of the zero polynomial is
(a) 0
(b) 1
(c) every real number
(d) not defined

#### Answer:

(d) not defined

Zero of the zero polynomial is not defined.

#### Question 14:

If p(x) = x = 4, then p(x) + p(−x) = ?
(a) 0
(b) 4
(c) 2x
(d) 8

#### Answer:

(d) 8

Let:
$p\left(x\right)=\left(x+4\right)$

Thus, we have:
$p\left(x\right)+p\left(-x\right)=\left\{\left(x+4\right)+\left(-x+4\right)\right\}$
= 4 + 4
=8

#### Question 15:

If $p\left(x\right)={x}^{2}-2\sqrt{2}x+1$, then $p\left(2\sqrt{2}\right)=$?
(a) 0
(b) 1
(c) $4\sqrt{2}$
(d) −1

#### Answer:

(b) 1

= 8 $-$ 8 + 1
= 1

#### Question 16:

The zeroes of the polynomial p(x) = x2 + x − 6 are
(a) 2, 3
(b) −2, 3
(c) 2, −3
(d) −2, −3

#### Answer:

(c) 2, −3

The zeroes of the polynomial p(x) is given by p(x) = 0.
We have:

Hence, 2 and $-$3 are the zeroes of the given polynomial.

#### Question 17:

The zeroes of the polynomial p(x) = 2x2 + 5x − 3 are
(a)
(b)
(c)
(d)

#### Answer:

(b)

The zeroes of the polynomial p(x) is given by p(x) = 0.
We have:

Hence,  are the zeroes of the given polynomial.

#### Question 18:

If (x2 + kx − 3) = (x − 3) (x + 1), then k = ?
(a) 2
(b) −2
(c) 3
(d) −1

#### Answer:

(b) −2

$\left(x-3\right)\left(x+1\right)={x}^{2}-2x-3$
Thus, we have:
${x}^{2}+kx-3={x}^{2}-2x-3\phantom{\rule{0ex}{0ex}}⇒k=-2$

#### Question 19:

If (x + 1) is a factor of (2x2 + kx), then k = ?
(a) −3
(b) −2
(c) 2
(d) 4

(c) 2

#### Question 20:

The coefficient of the highest power of x in the polynomial 2x3 − 4x4 + 5x2x5 + 3 is
(a) 2
(b) −4
(c) 3
(d) − 1

#### Answer:

(d) −1

The highest power of x in the given polynomial is 5.
The coefficient of x5 is −1.

#### Question 21:

When (x31 = 31) is divided by (x + 1), the remainder is
(a) 0
(b) 1
(c) 30
(d) 31

#### Answer:

(c) 30

Let:
$p\left(x\right)=\left({x}^{31}+31\right)\phantom{\rule{0ex}{0ex}}$
$x+1=0⇒x=-1$
By the remainder theorem, we know that when p(x) is divided by (x + 1), the remainder is p($-$1).
Thus, we have:

#### Question 22:

When p(x) = x3ax2 + a is divided by (xa), the remainder is
(a) 0
(b) a
(c) 2a
(d) 3a

#### Answer:

(b) a

$x-a=0⇒x=a$
By the remainder theorem, we know that when p(x) is divided by (x $-$ a), the remainder is p(a).
Thus, we have:

#### Question 23:

When p(x) = x3 + ax2 + 2x + a is divided by (x + a), the remainder is
(a) 0
(b) −1
(c) −15
(d) 21

#### Answer:

(c) −a

$x+a=0⇒x=-a$
By the remainder theorem, we know that when p (x) is divided by (x + a), the remainder is p (−a).
Thus, we have:

#### Question 24:

When p(x) = x4 + 2x3 − 3x2 + x − 1 is divided by (x − 2), the remainder is
(a) 0
(b) −1
(c) −15
(d) 21

#### Answer:

(d) 21

$x-2=0⇒x=2$
By the remainder theorem, we know that when p(x) is divided by (x $-$ 2), the remainder is p(2).
Thus, we have:

#### Question 25:

When p(x) = x3 − 3x2 + 4x + 32 is divided by (x + 2), the remainder is
(a) 0
(b) 32
(c) 36
(d) 4

#### Answer:

(d) 4

$x+2=0⇒x=-2$
By the remainder theorem, we know that when p(x) is divided by (x + 2), the remainder is p($-$2).
Now, we have:

#### Question 26:

When p(x) = 4x3 − 12x2 + 11x − 5 is divided by (2x − 1), the remainder is
(a) 0
(b) −5
(c) −2
(d) 2

#### Answer:

(c) −2

$2x-1=0⇒x=\frac{1}{2}$
By the remainder theorem, we know that when p(x) is divided by (2x $-$ 1), the remainder is $p\left(\frac{1}{2}\right)$.
Now, we have:

#### Question 27:

(x + 1) is a factor of the polynomial
(a) x3 − 2x2 + x + 2
(b) x3 + 2x2 + x − 2
(c) x3 − 2x2 − x − 2
(d) x3 − 2x2 − x + 2

#### Answer:

(c) x3 − 2x2 − x − 2

Let:
$f\left(x\right)={x}^{3}-2{x}^{2}+x+2$
By the factor theorem, (x + 1) will be a factor of f (x) if f (−1) = 0.
We have:

Hence, (x + 1) is not a factor of $f\left(x\right)={x}^{3}-2{x}^{2}+x+2$.

Now,
Let:
$f\left(x\right)={x}^{3}+2{x}^{2}+x-2$

By the factor theorem, (x + 1) will be a factor of f (x) if f ($-$1) = 0.
We have:

Hence, (x + 1) is not a factor of $f\left(x\right)={x}^{3}+2{x}^{2}+x-2$.

Now,
Let:
$f\left(x\right)={x}^{3}+2{x}^{2}-x-2$

By the factor theorem, (x + 1) will be a factor of f (x) if f ($-$1) = 0.
We have:

Hence, (x + 1) is a factor of $f\left(x\right)={x}^{3}+2{x}^{2}-x-2$.

#### Question 28:

(4x2 + 4x − 3) = ?
(a) (2x − 1) (2x − 3)
(b) (2x + 1) (2x − 3)
(c) (2x + 3) (2x − 1)
(d) none of these

#### Answer:

(c) (2x + 3) (2x − 1)

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

#### Question 29:

6x2 + 17x + 5 = ?
(a) (2x + 1)(3x + 5)
(b) (2x + 5)(3x + 1)
(c) (6x + 5)(x + 1)
(d) none of these

#### Answer:

(b) (2x + 5)(3x + 1)

$6{x}^{2}+17x+5=6{x}^{2}+15x+2x+5$
$=3x\left(2x+5\right)+1\left(2x+5\right)\phantom{\rule{0ex}{0ex}}=\left(2x+5\right)\left(3x+1\right)$

#### Question 30:

(x2 − 4x − 21) = ?
(a) (x − 7)(x − 3)
(b) (x + 7)(x − 3)
(c) (x − 7)(x + 3)
(d) none of these

#### Answer:

(c) (x − 7)(x + 3)

${x}^{2}-4x-21={x}^{2}-7x+3x-21$
$=x\left(x-7\right)+3\left(x-7\right)\phantom{\rule{0ex}{0ex}}=\left(x-7\right)\left(x+3\right)$

#### Question 31:

If (x + 5) is a factor of p(x) = x3 − 20x + 5k, then k = ?
(a) −5
(b) 5
(c) 3
(d) −3

(b) 5

#### Question 32:

3x3 + 2x2 + 3x + 2 = ?
(a) (3x − 2)(x2 − 1)
(b) (3x − 2)(x2 + 1)
(c) (3x + 2)(x2 − 1)
(d) (3x + 2)(x2 + 1)

#### Answer:

(d) (3x + 2)(x2 + 1)

$3{x}^{3}+2{x}^{2}+3x+2={x}^{2}\left(3x+2\right)+1\left(3x+2\right)$
$=\left(3x+2\right)\left({x}^{2}+1\right)$

#### Question 33:

If $\frac{x}{y}+\frac{y}{x}=-1$, where x ≠ 0 and y ≠ 0, then the value of (x3y3) is
(a) 1
(b) −1
(c) 0
(d) $\frac{1}{2}$

#### Answer:

(c) 0

$⇒x$2 + y2 = $-$xy
x2 + y2 + xy = 0

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

#### Question 34:

If a + b + c = 0, then a3 + b3 + c3 = ?
(a) 0
(b) abc
(c) 2abc
(d) 3abc

#### Answer:

(d) 3abc

$⇒{\left(a+b\right)}^{3}={\left(-c\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒{a}^{3}+{b}^{3}+3ab\left(a+b\right)=-{c}^{3}\phantom{\rule{0ex}{0ex}}⇒{a}^{3}+{b}^{3}+3ab\left(-c\right)=-{c}^{3}\phantom{\rule{0ex}{0ex}}⇒{a}^{3}+{b}^{3}+{c}^{3}=3abc$

#### Question 35:

If (x + 2) and (x − 1) are factors of (x3 + 10x2 + mx + n), then
(a) m = 5, n = −3
(b) m = 7, n = −18
(c) m = 17, n = −8
(d) m = 23, n = −19

#### Answer:

(b) m = 7, n = −18

Let:
$p\left(x\right)={x}^{3}+10{x}^{2}+mx+n\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Now,
$x+2=0⇒x=-2$
(x + 2) is a factor of p(x).
So, we have p($-$2)=0

Now,
$x-1=0⇒x=1$
Also,
(x $-$ 1) is a factor of p(x).
We have:
p(1) = 0

By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18

#### Question 36:

The value of (369)2 − (368)2 = ?
(a) 12
(b) 81
(c) 37
(d) 737

#### Answer:

(d) 737

${\left(369\right)}^{2}-{\left(368\right)}^{2}=\left(369+368\right)\left(369-368\right)$
$=\left(737×1\right)\phantom{\rule{0ex}{0ex}}=737$

104 × 96 = ?
(a) 9864
(b) 9984
(c) 9684
(d) 9884

#### Answer:

(b) 9984

$104×96=\left(100+4\right)\left(100-4\right)$
$={100}^{2}-{4}^{2}\phantom{\rule{0ex}{0ex}}=\left(10000-16\right)\phantom{\rule{0ex}{0ex}}=9984$

#### Question 38:

4a2 + b2 + 4ab + 8a + 4b + 4 = ?
(a) (2a + b + 2)2
(b) (2ab + 2)2
(c) (a + 2b + 2)2
(d) none of these

#### Answer:

(a) (2a + b + 2)2

#### Question 39:

The coefficient of x in the expansion of (x + 3)3 is
(a) 1
(b) 9
(c) 18
(d) 27

#### Answer:

(d) 27

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

Hence, the coefficient of x is 27.

#### Question 40:

If a + b + c = 0, then $\left(\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}\right)$=?

#### Answer:

(d) 3

$a+b+c=0⇒{a}^{3}+{b}^{3}+{c}^{3}=3abc$

Thus, we have:
$\left(\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}\right)=\frac{{a}^{3}+{b}^{3}+{c}^{3}}{abc}$
$=\frac{3abc}{abc}\phantom{\rule{0ex}{0ex}}=3$

#### Question 41:

If x + y + z = 9 and xy + yz + zx = 23, then the value of (x3 + y3 + z3 − 3xyz) = ?
(a) 108
(b) 207
(c) 669
(d) 729

#### Answer:

(a) 108

${x}^{3}+{y}^{3}+{z}^{3}-3xyz=\left(x+y+z\right)\left({x}^{2}+{y}^{2}+{z}^{2}-xy-yz-zx\right)\phantom{\rule{0ex}{0ex}}$
$=\left(x+y+z\right)\left[{\left(x+y+z\right)}^{2}-3\left(xy+yz+zx\right)\right]\phantom{\rule{0ex}{0ex}}=9×\left(81-3×23\right)\phantom{\rule{0ex}{0ex}}=9×12\phantom{\rule{0ex}{0ex}}=108$

#### Question 42:

If (x100 + 2x99 + k) is divisible by (x + 1), then the value of k is
(a) 1
(b) 2
(c) −2
(d) −3

#### Answer:

(a) 1

Let:
$p\left(x\right)={x}^{100}+2{x}^{99}+k$
Now,
$x+1=0⇒x=-1$

#### Question 43:

In a polynomial in x, the indices of x must be
(a) integers
(b) positive integers
(c) non-negative integers
(d) real numbers

#### Answer:

(c) non-negative integers

#### Question 44:

For what value of k is the polynomial p(x) = 2x3kx + 3x + 10 exactly divisible by (x + 2)?
(a) $-\frac{1}{3}$
(b) $\frac{1}{3}$
(c) 3
(d) −3

#### Answer:

(d) −3

Let:
$p\left(x\right)=2{x}^{3}-k{x}^{2}+3x+10\phantom{\rule{0ex}{0ex}}$
Now,
$x+2=0⇒x=-2$

207 × 193 = ?
(a) 39851
(b) 39951
(c) 39961
(d) 38951

(b) 39951

305 × 308 = ?
(a) 94940
(b) 93840
(c) 93940
(d) 94840

(c) 93940

#### Question 47:

The zeroes of the polynomial p(x) = x2 − 3x are
(a) 0, 0
(b) 0, 3
(c) 0, −3
(d) 3, −3

#### Answer:

(b) 0, 3

Let:
$p\left(x\right)={x}^{2}-3x$
Now, we have:
$p\left(x\right)=0⇒{x}^{2}-3x=0$

#### Question 48:

The zeroes of the polynomial p(x) = 3x2 − 1 are
(a) $\frac{1}{3}$
(b) $\frac{1}{\sqrt{3}}$
(c) $\frac{-1}{\sqrt{3}}$
(d) $\frac{1}{\sqrt{3}}\mathrm{and}\frac{-1}{\sqrt{3}}$

#### Answer:

(d)

Let:
$p\left(x\right)=3{x}^{2}-1$

#### Question 49:

Assertion: If (x − 1) is a factor of p(x) = x2 + kx + 1, then k = −2.
Reason: If (xa) is a factor of p(x), then p(a) = 0.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and but Reason is not a correct explanation of assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

#### Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Assertion:
Let:
$p\left(x\right)={x}^{2}+kx+1\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Now,
$x-1=0⇒x=1$

Hence, Assertion is true.
Reason: If (xa) is a factor of p(x), then p(a) = 0.
The given statement is true.
Therefore, both Assertion and Reason are true and Reason is a correct explanation of Assertion.

#### Question 50:

Assertion: If p(x) = x3ax2 + 6xa is divided by (xa), then the remainder is 5a.
Reason: If p(x) is divided by (xa),then the remainder is p(a).

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

#### Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Assertion:
Let:
$p\left(x\right)={x}^{3}-a{x}^{2}+6x-a$
Now,
$x-a=0⇒x=a$
By the remainder theorem, we know that when
Now, we have:

Hence, Assertion is true.
Reason: If p(x) is divided by (xa), then the remainder is p(a).
The given statement is true.
Therefore, both Assertion and Reason are true and Reason is a correct explanation of Assertion.

#### Question 51:

Assertion: If (x − 2) is a factor p(x) = x3 − 2x + 3k, then $k=\frac{-4}{3}$.
Reason: If p(x) is divided by (xa), then the remainder is p(a).

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

#### Answer:

(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.

Assertion:
Let:
$p\left(x\right)={x}^{3}-2x+3k$
Now,
$x-2=0⇒x=2$

Hence, Assertion is true.
Reason: If p(x) is divided by (xa), then the remainder is p(a).
The given statement is true, but it is not correct explanation of Assertion.
Therefore, both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.

#### Question 52:

Assertion: The value of (25)3 + (−16)3 + (−9)3 is 10800.
Reason: If a + b + c = 0, then a3 + b3 + c3 = 3abc.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true and but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

#### Answer:

(a) Both Assertion and Reason are true, and Reason is a correct explanation of Assertion.

Assertion:
Let:

Now,
$a+b+c=25+\left(-16\right)+\left(-9\right)\phantom{\rule{0ex}{0ex}}$
= 0
$\because a+b+c=0\phantom{\rule{0ex}{0ex}}\therefore {a}^{3}+{b}^{3}+{c}^{3}=3abc$

Thus, we have:
$3abc=\left[3×25×\left(-16\right)×\left(-9\right)\right]$
= 10800
Hence, Assertion is true.
Reason: If a + b + c = 0, then a3 + b3 + c3 = 3abc.
The given statement is true.
Therefore,  both Assertion and Reason are true and Reason is a correct explanation of Assertion.

#### Question 53:

Match of following columns:

 Column I Column II (a) When p(x) = 81x4 + 54x3 − 9x2 − 3x + 2 is divided by (3x + 2), then reaminder ....... . (p) 1 (b) p(x) = ax3 + 3x2 − 3 and q(x) = 2x3 − 5x + a divided by (x − 4) leave the same remainder. Then, a ...... . (q) −7 (c) If $p\left(x\right)=7{x}^{2}-4\sqrt{2}x+c$is completely divisible by $\left(x-\sqrt{2}\right)$, then c = ...... . (r) 0 (d) If q(x) = 2x3 + bx2 + 11x + b + 3 is divisibl by (2x + 1), then b = ....... . (s) −6 Column II (a) If p(x) = x3 + 2x2 + 3x + 1 is divided by (x + 1), then remainder = ...... . (p) $\frac{7}{8}$ (b) If (2x − 1) is a factor of q(x) = x3 + 2x2 − 3x + k, then k = ..... . (q) 50 (c) The degree of the constant polynomial (−5) is = ....... . (r) −1 (d) When x51 + 51 is divided by (x + 1), the remainder is = ...... . (s) 0

(a) ...... .
(b) ...... .
(c) ...... .
(d) ...... .

#### Answer:

(a)
Let:
$p\left(x\right)={x}^{3}+2{x}^{2}+3x+1$
$x+1=0⇒x=-1$
By the remainder theorem, we know that when
Thus, we have:

∴ $\left(a\right)⇒\left(r\right)$

(b)
Let:
$q\left(x\right)={x}^{3}+2{x}^{2}-3x+k$
Now,
$2x-1=0⇒x=\frac{1}{2}$

∴ $\left(b\right)⇒\left(p\right)$

(c)
The degree of the constant polynomial ($-$5) is 0.
∴ $\left(c\right)⇒\left(s\right)$

(d)
Let:
$p\left(x\right)={x}^{51}+51$
Now,
$x+1=0⇒x=-1$
By the remainder theorem, we know that when
Thus, we have:
$p\left(-1\right)={\left(-1\right)}^{51}+51$
= $-$1+51
= 50
∴ $\left(d\right)⇒\left(q\right)$

#### Question 54:

Match of the following columns:

 Column I Column II (a) When p(x) = 81x4 + 54x3 − 9x2 − 3x + 2 is divided by (3x + 2), then reaminder ....... . (p) 1 (b) p(x) = ax3 + 3x2 − 3 and q(x) = 2x3 − 5x + a divided by (x − 4) leave the same remainder. Then, a ...... . (q) −7 (c) If $p\left(x\right)=7{x}^{2}-4\sqrt{2}x+c$is completely divisible by $\left(x-\sqrt{2}\right)$, then c = ...... . (r) 0 (d) If q(x) = 2x3 + bx2 + 11x + b + 3 is divisibl by (2x + 1), then b = ....... . (s) −6

(a) ....... .
(b) ....... .
(c) ....... .
(d) ....... .

#### Answer:

(a)
Let:
$p\left(x\right)=81{x}^{4}+54{x}^{3}-9{x}^{2}-3x+2$
$3x+2=0⇒x=\frac{-2}{3}$
By the remainder theorem, we know that when
Thus, we have:
$p\left(\frac{-2}{3}\right)=81×{\left(\frac{-2}{3}\right)}^{4}+54×{\left(\frac{-2}{3}\right)}^{3}-9×{\left(\frac{-2}{3}\right)}^{2}-3×\left(\frac{-2}{3}\right)+2$
$=\left(81×\frac{16}{81}\right)-\left(54×\frac{8}{27}\right)-\left(9×\frac{4}{9}\right)+2+2\phantom{\rule{0ex}{0ex}}=\left(16-16-4+2+2\right)\phantom{\rule{0ex}{0ex}}=0$
∴ $\left(a\right)⇒\left(r\right)$

(b)
Let:
$p\left(x\right)=a{x}^{3}+3{x}^{2}-3$
$x-4=0⇒x=4$
By the remainder theorem, we know that when
Thus, we have:

Also,
Let:
$q\left(x\right)=2{x}^{3}-5x+a$
By the remainder theorem, we know that when​
Thus, we have:

Now,
$64a+45=108+a\phantom{\rule{0ex}{0ex}}⇒63a=63\phantom{\rule{0ex}{0ex}}⇒a=1$
∴ $\left(b\right)⇒\left(p\right)$

(c)
Let:
$p\left(x\right)=7{x}^{2}+-4\sqrt{2}x+c$
Now,
$x-\sqrt{2}=0⇒x=\sqrt{2}$

$\left(c\right)⇒\left(s\right)$

(d)
Let:
$q\left(x\right)=2{x}^{3}+b{x}^{2}+11x+b+3$
Now,
$2x-1=0⇒x=\frac{1}{2}$

$⇒\frac{1+b+22+4b+12}{4}=0\phantom{\rule{0ex}{0ex}}⇒35+5b=0\phantom{\rule{0ex}{0ex}}⇒b=\frac{-35}{5}\phantom{\rule{0ex}{0ex}}⇒b=-7$
∴ $\left(d\right)⇒\left(q\right)$

#### Question 1:

Let p(x) = 3x3 + 4x2 − 5x + 8. Find p(−2).

#### Answer:

Let:
$p\left(x\right)=3{x}^{3}+4{x}^{2}-5x+8$
Thus, we have:
$p\left(-2\right)=3×{\left(-2\right)}^{3}+4×{\left(-2\right)}^{2}-5×\left(-2\right)+8$

#### Question 2:

Find the remainder when p(x) = 4x3 + 8x2 − 17x + 10 is divided by (2x − 1).

#### Answer:

Let:
$p\left(x\right)=4{x}^{3}+8{x}^{2}-17x+10$
$\mathrm{And},\phantom{\rule{0ex}{0ex}}2x-1=0⇒x=\frac{1}{2}$
By the remainder theorem, we know that when
Thus, we have:
$p\left(\frac{1}{2}\right)=4×{\left(\frac{1}{2}\right)}^{3}+8×{\left(\frac{1}{2}\right)}^{2}-17×\frac{1}{2}+10$
$=\frac{1}{2}+2-\frac{17}{2}+10\phantom{\rule{0ex}{0ex}}=4$

∴ Remainder = 4

#### Question 3:

If (x − 2) is a factor of 2x3 − 7x2 + 11x + 5a, find the value of a.

#### Answer:

Let:
$p\left(x\right)=2{x}^{3}-7{x}^{2}+11x+5a$
Now, we have:
$x-2=0⇒x=2$

Hence, the value of a is $-$2.

#### Question 4:

For what value of m, p(x) = (x3 − 2mx2 + 16) is divisible by (x + 2)?

#### Answer:

$\mathrm{Let}:\phantom{\rule{0ex}{0ex}}p\left(x\right)={x}^{3}-2m{x}^{2}+16$
Now, we have:
$x+2=0⇒x=-2$

Hence, the value of m is 1.

#### Question 5:

If (a + b + c) = 8 and (ab + bc + ca) = 19, find (a2 + b2 + c2).

#### Answer:

$={8}^{2}-\left(2×19\right)\phantom{\rule{0ex}{0ex}}=64-38\phantom{\rule{0ex}{0ex}}=26$

#### Question 6:

Expand: (3a + 4b + 5c)2.

#### Answer:

${\left(3a+4b+5c\right)}^{2}={\left(3a\right)}^{2}+{\left(4b\right)}^{2}+{\left(5c\right)}^{2}+2\left(3a\right)\left(4b\right)+2\left(4b\right)\left(5c\right)+2\left(3a\right)\left(5c\right)$
$=9{a}^{2}+16{b}^{2}+25{c}^{2}+24ab+40bc+30ac$

#### Question 7:

Expand: (3x + 2)3.

#### Answer:

${\left(3x+2\right)}^{3}={\left(3x\right)}^{3}+{2}^{3}+\left(3×3x×2\right)×\left(3x+2\right)$
$=27{x}^{3}+8+18x×\left(3x+2\right)\phantom{\rule{0ex}{0ex}}=27{x}^{3}+8+54{x}^{2}+36x\phantom{\rule{0ex}{0ex}}=27{x}^{3}+54{x}^{2}+36x+8$

#### Question 8:

Evaluate: {(28)3 + (−15)3 + (−13)3}.

#### Answer:

Let:

Now, we have:
$a+b+c=28+\left(-15\right)+\left(-13\right)$
= 0

#### Question 9:

If (x60 + 60) is divided by (x + 1), the remainder is

(a) 0
(b) 59
(c) 61
(d) 2

#### Answer:

(c) 61

Let:
$p\left(x\right)={x}^{60}+60$
$x+1=0⇒x=-1$
By the remainder theorem, we know that when
Now, we have:

#### Question 10:

One of the factors of (36x2 − 1) + (1 + 6x)2 is

(a) (6x − 1)
(b) (6x + 1)
(c) 6x
(d) 6 − x

#### Answer:

(b) (6x + 1)

$\left(36{x}^{2}-1\right)+{\left(1+6x\right)}^{2}=\left[{\left(6x\right)}^{2}-{1}^{2}\right]+{\left(1+6x\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
$=\left(6x+1\right)\left(6x-1\right)+{\left(6x+1\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(6x+1\right)\left[\left(6x-1\right)+1\right]\phantom{\rule{0ex}{0ex}}=\left(6x+1\right)6x$

Hence, one of the factors of the given polynomial is (6x + 1).

#### Question 11:

If $\frac{a}{b}+\frac{b}{a}=-1$, then (a3b3) = ?

(a) −1
(b) −3
(c) −2
(d) 0

#### Answer:

(d) 0

$⇒{a}^{2}+{b}^{2}+ab=0$

Thus, we have:
${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$
$=\left(a-b\right)×0=0$

#### Question 12:

The coefficient of x in the expansion of (x + 5)3 is

(a) 1
(b) 15
(c) 45
(d) 75

#### Answer:

(d) 75

${\left(x+5\right)}^{3}={x}^{3}+{5}^{3}+\left(3×x×5\right)\left(x+5\right)$
$={x}^{3}+125+15x×\left(x+5\right)\phantom{\rule{0ex}{0ex}}={x}^{3}+125+15{x}^{2}+75x$

Hence, the coefficient of x is 75.

#### Question 13:

$\sqrt{5}$is a polynomial of degree

(a) $\frac{1}{2}$
(b) 2
(c) 0
(d) 1

#### Answer:

(c) 0

$\sqrt{5}$ is a non-zero constant term and a non-zero constant is a polynomial of degree 0.

#### Question 14:

One of the zeroes of the polynomial 2x2 + 7x − 4 is

(a) 2
(b) $\frac{1}{2}$
(c) −2
(d) $\frac{-1}{2}$

#### Answer:

(b) $\frac{1}{2}$

Let:
$p\left(x\right)=2{x}^{2}+7x-4$
$p\left(x\right)=0⇒2{x}^{2}+7x-4=0$

Hence, one of the zeroes of the given polynomial is $\frac{1}{2}$.

#### Question 15:

Zero of the zero polynomial is

(a) 0
(b) 1
(c) every real number
(d) not defined

#### Answer:

(d) not defined

Zero of the zero polynomial is not defined.

#### Question 16:

If (x + 1) and (x − 1) are factors of p(x) = ax3 + x2 − 2x + b, find the values of a and b.

#### Answer:

Let:
$p\left(x\right)=a{x}^{3}+{x}^{2}-2x+b$
Now,
$x=1=0⇒x=-1$

Also,
$x-1=0⇒x=1$

By substituting the value of a, we get the value of b, i.e., −1.
∴ a = 2 and b = −1

#### Question 17:

If (x + 2) is a factor of p(x) = ax3 + bx2 + x − 6 and p(x) when divided by (x − 2) leaves a remainder 4, prove that a = 0 and b = 2.

#### Answer:

Let:
$p\left(x\right)=a{x}^{3}+b{x}^{2}+x-6$
Now,

Also,
$x-2=0⇒x=2$
When p(x) is divided by (x $-$ 2), the remainder is 4.
p(2) = 4

Thus, we have:

By substituting the value of a, we get the value of b, i.e., 2.
∴ a = 0 and b = 2

#### Question 18:

The expanded form of (3x − 5)3 is

(a) 27x3 + 135x2 + 225x − 125
(b) 27x3 + 135x2 − 225x − 125
(c) 27x3 − 135x2 + 225x − 125
(d) none of these

#### Answer:

(c) 27x3 − 135x2 + 225x − 125

${\left(3x-5\right)}^{3}={\left(3x\right)}^{3}-{5}^{3}-\left(3×3x×5\right)\left(3x-5\right)$
$=27{x}^{3}-125-45x\left(3x-5\right)\phantom{\rule{0ex}{0ex}}=27{x}^{3}-125-135{x}^{2}+225x\phantom{\rule{0ex}{0ex}}=27{x}^{3}-135{x}^{2}+225x-125$

#### Question 19:

If a + b + c = 5 and ab + bc + ca = 10, prove that a3 + b3 + c3 − 3abc = −25.

#### Answer:

Now,

$=5×\left(5-10\right)\phantom{\rule{0ex}{0ex}}=5×\left(-5\right)\phantom{\rule{0ex}{0ex}}=-25$

#### Question 20:

If p(x) = 2x3 + ax2 + 3x − 5 and q(x) = 2x3 + ax2 + 3x − 5 leave the same remainder when divided by (x − 2), show that $a=\frac{-13}{3}$.

#### Answer:

Let:
$p\left(x\right)=2{x}^{3}+a{x}^{2}+3x-5$
Now,
$x-2=0⇒x=2$
When p(x) is divided by (x $-$ 2), then the remainder is p(2).

Let:
$q\left(x\right)={x}^{3}+{x}^{2}-4x+a$
When q(x) is divided by (x $-$ 2), then the remainder is q(2).

According to the question, p(2) and q(2) are the same.
Thus, we have:
$4a+17=a+4\phantom{\rule{0ex}{0ex}}⇒3a=-13\phantom{\rule{0ex}{0ex}}⇒a=\frac{-13}{3}$

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