Exercise 2.3 NCERT Class 9 Chapter 2 Polynomials Best Solutions

Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 Exercise 2.1 Exercise 2.2 Exercise 2.3 Exercise 2.4

Exercise 2.3 NCERT Class 9 contains a total of five questions. All questions are based on the topic: Factorisation of Polynomials and Factor Theorem.

Factor Theorem:

For a polynomial $p (x)$ and $(x - a)$
$(i)$ If $p (a) = 0$ then $(x - a)$ is a factor of $p (x)$ .
$(i i)$ If $(x - a)$ is a factor of $p (x)$ then $p (a) = 0$ .

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Exercise 2.3 NCERT Class 9 Polynomials Mathematics Solutions

Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7 Q.8 Q.9

The first question of Exercise 2.3 NCERT Class 9 is about determining a polynomial to be a factor of another polynomial using factor theorem.

1. Determine which of the following polynomials has $(x + 1)$ a factor:

(i) $x^{3} + x^{2} + x + 1$

Answer

Let $p (x) = x^{3} + x^{2} + x + 1$ and $g (x) = x + 1$ .

Here, $g (x) = 0$ $\Rightarrow x + 1 = 0$ $\Rightarrow x = - 1$

According to the factor theorem, if $p (- 1) = 0$ , then $(x + 1)$ is a factor of $p (x)$

$p (- 1) = {(- 1)}^{3} + {(- 1)}^{2} + (- 1) + 1$
$= - 1 + 1 - 1 + 1$
$= 0$

$∵ p (- 1) = 0,$ $(x + 1)$ is a factor of $p (x)$ .

(ii) $x^{4} + x^{3} + x^{2} + x + 1$

Answer

Let $p (x) = x^{4} + x^{3} + x^{2} + x + 1$ and $g (x) = x + 1$ .

Here, $g (x) = 0$ $\Rightarrow x + 1 = 0$ $\Rightarrow x = - 1$

According to the factor theorem, if $p (- 1) = 0$ , then $(x + 1)$ is a factor of $p (x)$

$p (- 1) = {(- 1)}^{4} + {(- 1)}^{3} + {(- 1)}^{2} + (- 1) + 1$
$= 1 - 1 + 1 - 1 + 1$
$= 1$

$∵ p (- 1) \neq 0$ $(x + 1)$ is not a factor of $p (x)$ .

(iii) $x^{4} + 3 x^{3} + 3 x^{2} + x + 1$

Answer

Let $p (x) = x^{4} + 3 x^{3} + 3 x^{2} + x + 1$ and $g (x) = x + 1$ .

Here, $g (x) = 0$ $\Rightarrow x + 1 = 0$ $\Rightarrow x = - 1$

According to the factor theorem, if $p (- 1) = 0$ , then $(x + 1)$ is a factor of $p (x)$

$p (- 1) = {(- 1)}^{4} + 3 {(- 1)}^{3} + 3 {(- 1)}^{2} + (- 1) + 1$
$= 1 - 3 + 3 - 1 + 1$
$= 1$

$∵ p (- 1) \neq 0$ $(x + 1)$ is not a factor of $p (x)$ .

(iv) $x^{3} - x^{2} - (2 + \sqrt{2}) x + \sqrt{2}$

Answer

Let $p (x) = x^{3} - x^{2} - (2 + \sqrt{2}) x + \sqrt{2}$ and $g (x) = x + 1$ .

Here, $g (x) = 0$ $\Rightarrow x + 1 = 0$ $\Rightarrow x = - 1$

According to the factor theorem, if $p (- 1) = 0$ , then $(x + 1)$ is a factor of $p (x)$

$p (- 1) = {(- 1)}^{3} - {(- 1)}^{2} - (2 + \sqrt{2}) (- 1) + \sqrt{2}$
$= - 1 - 1 + (2 + \sqrt{2}) + \sqrt{2}$
$= - 2 + 2 + \sqrt{2} + \sqrt{2}$
$= 2 \sqrt{2}$

$∵ p (- 1) \neq 0$ $(x + 1)$ is not a factor of $p (x)$ .

The second question of Exercise 2.3 NCERT Class 9 is wholly about using the factor theorem.

2. Use the Factor Theorem to determine whether $g (x)$ is a factor of $p (x)$ in each of the following cases:

(i) $p (x) = 2 x^{3} + x^{2} - 2 x - 1,$ $g (x) = x + 1$

Answer

Given that
$p (x) = 2 x^{3} + x^{2} - 2 x - 1,$
$g (x) = x + 1$

Here, $g (x) = 0$ $\Rightarrow x + 1 = 0$ $\Rightarrow x = - 1$

According to the factor theorem, if $p (- 1) = 0$ , then $(x + 1)$ is a factor of $p (x)$

$p (- 1) = 2 {(- 1)}^{3} + {(- 1)}^{2} - 2 (- 1) - 1$
$= - 2 + 1 + 2 - 1$
$= - 3 + 3$
$= 0$

$∵ p (- 1) = 0,$ $(x + 1)$ is a factor of $p (x)$ .

(ii) $p (x) = x^{3} + 3 x^{2} + 3 x + 1,$ $g (x) = x + 2$

Answer

Given that
$p (x) = x^{3} + 3 x^{2} + 3 x + 1,$
$g (x) = x + 2$

Here, $g (x) = 0$ $\Rightarrow x + 2 = 0$ $\Rightarrow x = - 2$

According to the factor theorem, if $p (- 2) = 0$ , then $(x + 2)$ is a factor of $p (x)$

$p (- 2) = {(- 2)}^{3} + 3 {(- 2)}^{2} + 3 (- 2) + 1$
$= - 8 + 12 - 6 + 1$
$= - 14 + 13$
$= - 1$

$∵ p (- 2) \neq 0$ , $(x + 2)$ is not a factor of $p (x)$ .

(iii) $p (x) = x^{3} - 4 x^{2} + x + 6,$ $g (x) = x - 3$

Answer

Given that
$p (x) = x^{3} - 4 x^{2} + x + 6,$
$g (x) = x - 3$

Here, $g (x) = 0$ $\Rightarrow x - 3 = 0$ $\Rightarrow x = 3$ .

According to the factor theorem, if $p (3) = 0$ , then $(x - 3)$ is a factor of $p (x)$ .

$p (3) = 3^{3} - 4 \times 3^{2} + 3 + 6$
$= 27 - 36 + 3 + 6$
$= 36 - 36$
$= 0$

$∵ p (3) = 0,$ $(x - 3)$ is a factor of $p (x)$ .

The third question of Exercise 2.3 NCERT Class 9 is about finding the value of a variable using factor theorem.

3. Find the value of $k$ , if $(x - 1)$ is a factor of $p (x)$ in each of the following cases:.

(i) $p (x) = x^{2} + x + k$

Answer

We have
$p (x) = x^{2} + x + k$

$∵ (x - 1)$ is a factor of $p (x)$ , so $p (1) = 0$
[factor theorem]

$∴ p (1) = 0$
$\Rightarrow 1^{2} + 1 + k = 0$
$\Rightarrow 1 + 1 + k = 0$
$\Rightarrow 2 + k = 0$
$\Rightarrow k = - 2$

Hence, the value of $k = - 2$ .

(ii) $p (x) = 2 x^{2} + k x + \sqrt{2}$

Answer

We have
$p (x) = 2 x^{2} + k x + \sqrt{2}$

$∵ (x - 1)$ is a factor of $p (x)$ , so $p (1) = 0$
[factor theorem]

$∴ p (1) = 0$
$\Rightarrow 2 \times 1^{2} + k \times 1 + \sqrt{2} = 0$
$\Rightarrow 2 + k + \sqrt{2} = 0$
$\Rightarrow k = - 2 - \sqrt{2}$
$\Rightarrow k = - (2 + \sqrt{2})$

Hence, $k = - (2 + \sqrt{2})$ .

(iii) $p (x) = k x^{2} - \sqrt{2} x + 1$

Answer

We have
$p (x) = k x^{2} - \sqrt{2} x + 1$

$∵ (x - 1)$ is a factor of $p (x)$ , so $p (1) = 0$
[factor theorem]

$∴ p (1) = 0$
$\Rightarrow k \times 1^{2} - \sqrt{2} \times 1 + 1 = 0$
$\Rightarrow k - \sqrt{2} + 1 = 0$
$\Rightarrow k = \sqrt{2} - 1$

Hence, $k = \sqrt{2} - 1$ .

(iv) $p (x) = k x^{2} - 3 x + k$

Answer

We have
$p (x) = k x^{2} - 3 x + k$

$∵ (x - 1)$ is a factor of $p (x)$ , so $p (1) = 0$
[factor theorem]

$∴ p (1) = 0$
$\Rightarrow k \times 1^{2} - 3 \times 1 + k = 0$
$\Rightarrow k - 3 + k = 0$
$\Rightarrow 2 k - 3 = 0$
$\Rightarrow 2 k = 3$
$\Rightarrow k = \frac{3}{2}$

Hence, $k = \frac{3}{2}$ .

The fourth question of Exercise 2.3 NCERT Class 9 is about factorising the given quadratic polynomial.

4. Factorise:

(i) $12 x^{2} - 7 x + 1$

Answer

Let $p (x) = 12 x^{2} - 7 x + 1$ .

Here, the coefficient of $x^{2}$ × constant term = 12 × 1 = 12.

$12 = 2 \times 2 \times 3 \times 1$ $= (2 \times 2) \times (3 \times 1)$ $= 4 \times 3$

Now,

$p (x) = 12 x^{2} - 7 x + 1$
$= 12 x^{2} - (4 + 3) x + 1$
$= 12 x^{2} - 4 x - 3 x + 1$
$= 4 x (3 x - 1) - 1 (3 x - 1)$
$= (3 x - 1) (4 x - 1)$

Hence, $12 x^{2} - 7 x + 1$ = $(3 x - 1) (4 x - 1)$

(ii) $2 x^{2} + 7 x + 3$

Answer

Let $p (x) = 2 x^{2} + 7 x + 3$

Here, the coefficient of $x^{2}$ × constant term = 2 × 3 = 6.

$6 = 2 \times 3 \times 1 = (2 \times 3) \times 1 = 6 \times 1$

$N$ ow,

$p (x) = 2 x^{2} + 7 x + 3$
$= 2 x^{2} + (6 + 1) x + 3$
$= 2 x^{2} + 6 x + x + 3$
$= 2 x (x + 3) + 1 (x + 3)$
$= (x + 3) (2 x + 1)$

Hence, $2 x^{2} + 7 x + 3$ = $(x + 3) (2 x + 1)$

(iii) $6 x^{2} + 5 x - 6$

Answer

Let $p (x) = 6 x^{2} + 5 x - 6$

Here, the coefficient of $x^{2}$ × constant term = 6 × (−6) = −36.

$36 = 2 \times 2 \times 3 \times 3 \times 1 = (2 \times 2) \times (3 \times 3 \times 1)$ $= 4 \times 9$

$N$ ow,

$p (x) = 6 x^{2} + 5 x - 6$
$= 6 x^{2} + (9 - 4) x - 6$
$= 6 x^{2} + 9 x - 4 x - 6$
$= 3 x (2 x + 3) - 2 (2 x + 3)$
$= (2 x + 3) (3 x - 2)$

Hence, $6 x^{2} + 5 x - 6$ = $(2 x + 3) (3 x - 2)$

(iv) $3 x^{2} - x - 4$

Answer

Let $p (x) = 3 x^{2} - x - 4$

Here, the coefficient of $x^{2}$ × constant term = 3 × (−4) = −12.

$12 = 2 \times 2 \times 3 \times 1 = (2 \times 2) \times (3 \times 1) = 4 \times 3$

$N$ ow,

$p (x) = 3 x^{2} - x - 4$
$= 3 x^{2} - (4 - 3) x - 4$
$= 3 x^{2} - 4 x + 3 x - 4$
$= x (3 x - 4) + 1 (3 x - 4)$
$= (x + 1) (3 x - 4)$

Hence, $3 x^{2} - x - 4$ = $(x + 1) (3 x - 4)$

The fifth question of Exercise 2.3 NCERT Class 9 is about factorisation of cubic polynomials.

5. Factorise:

(i) $x^{3} - 2 x^{2} - x + 2$

Answer

Let $p (x) = x^{3} - 2 x^{2} - x + 2$

We shall now look for all the factors of $2$ .
These are $\pm 1, \pm 2$ .

Now, we find the values of $p (1), p (2), p (- 1), p (- 2)$ and we determine if any of these values are zero.

$∵ p (1) = 1^{3} - 2 \times 1^{2} - 1 + 2$
$= 1 - 2 - 1 + 2$
$= 3 - 3$
$= 0$

$∴ (x - 1)$ is a factor of $p (x)$ .

Now, we divide $p (x)$ by $(x - 1)$ to get a quadratic polynomial as quotient.

So, $x^{3} - 2 x^{2} - x + 2$ $= (x - 1) (x^{2} - x - 2)$
$= (x - 1) (x^{2} - 2 x + x - 2)$
$= (x - 1) [x (x - 2) + 1 (x - 2)]$
$= (x - 1) (x + 1) (x - 2)$

Or, we can do the following to easily obtain the quadratic polynomial.

$x^{3} - 2 x^{2} - x + 2$

$= x^{3} - x^{2} - x^{2} + x - x - x + 2$ ... $[∵ - 2 x^{2} = - x^{2} - x^{2}, x - x = 0]$

$= x^{3} - x^{2} - x^{2} + x - 2 x + 2$

$= (x^{3} - x^{2}) - (x^{2} - x) - (2 x - 2)$

$= x^{2} (x - 1) - x (x - 1) - 2 (x - 1)$

$= (x - 1) (x^{2} - x - 2)$

$= (x - 1) (x^{2} - 2 x + x - 2)$

$= (x - 1) [x (x - 2) + 1 (x - 2)]$

$= (x - 1) (x + 1) (x - 2)$ .

Hence, $x^{3} - 2 x^{2} - x + 2$ $= (x - 1) (x + 1) (x - 2)$ .

(ii) $x^{3} - 3 x^{2} - 9 x - 5$

Answer

Let $p (x) =$ $x^{3} - 3 x^{2} - 9 x - 5$

We shall now look for all the factors of $- 5$ .
These are $\pm 1, \pm 5$ .

Now, we find the values of $p (1), p (5), p (- 1), p (- 5)$ and we determine if any of these values are zero.

$∵ p (- 1) = {(- 1)}^{3} - 3 {(- 1)}^{2} - 9 (- 1) - 5$
$= - 1 - 3 + 9 - 5$
$= - 9 + 9$
$= 0$

$∴ (x + 1)$ is a factor of $p (x)$ .

Now, we divide $p (x)$ by $(x + 1)$ to get a quadratic polynomial as quotient.

So, $x^{3} - 3 x^{2} - 9 x - 5$ $= (x + 1) (x^{2} - 4 x - 5)$
$= (x + 1) (x^{2} - 5 x + x - 5)$
$= (x + 1) [x (x - 5) + 1 (x - 5)]$
$= (x + 1) (x + 1) (x - 5)$

Or, we can also do the following to easily obtain the factors.

$x^{3} - 3 x^{2} - 9 x - 5$

$= x^{3} + x^{2} - x^{2} - 3 x^{2} - 9 x - 5$ ... $[∵ x^{2} - x^{2} = 0]$

$= x^{3} + x^{2} - 4 x^{2} - 9 x - 5$ ... $[∵ - x^{2} - 3 x^{2} = 4 x^{2}]$

$= x^{3} + x^{2} - 4 x^{2} - 9 x - 5$

$= x^{3} + x^{2} - 4 x^{2} - 4 x - 5 x - 5$ ... $[∵ - 9 x = - 4 x - 5 x]$

$= (x^{3} + x^{2}) - (4 x^{2} + 4 x) - (5 x + 5)$

$= x^{2} (x + 1) - 4 x (x + 1) - 5 (x + 1)$

$= (x + 1) (x^{2} - 4 x - 5)$

$= (x + 1) (x + 1) (x - 5)$ ... $[∵ x^{2} - 4 x - 5 = (x + 1) (x - 5)]$

Hence, $x^{3} - 3 x^{2} - 9 x - 5$ $= (x + 1) (x + 1) (x - 5)$ .

(iii) $x^{3} + 13 x^{2} + 32 x + 20$

Answer

Let $p (x)$ = $x^{3} + 13 x^{2} + 32 x + 20$

We shall now look for all the factors of $20$ .
These are $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10$ .

Now, we find the values of $p (1), p (- 1), p (2), p (- 2), p (4), p (- 4),$ $p (5), p (- 5), p (10), p (- 10)$ and we determine if any of these values are zero.

$∵ p (- 1) = {(- 1)}^{3} + 13 {(- 1)}^{2} + 32 (- 1) + 20$
$= - 1 + 13 - 32 + 20$
$= - 33 + 33$
$= 0$

$∴ (x + 1)$ is a factor of $p (x)$ .

Now, we divide $p (x)$ by $(x + 1)$ to get a quadratic polynomial as quotient.

So, $x^{3} + 13 x^{2} + 32 x + 20$ $= (x + 1) (x^{2} + 12 x + 20)$
$= (x + 1) (x^{2} + 10 x + 2 x + 20)$
$= (x + 1) [x (x + 10) + 2 (x + 10)]$
$= (x + 1) (x + 2) (x + 10)$

Or, we can also do the following to easily obtain the factors.

$x^{3} + 13 x^{2} + 32 x + 20$

$= x^{3} + x^{2} + 12 x^{2} + 32 x + 20$ ... $[∵ 13 x^{2} = x^{2} + 12 x^{2}]$

$= x^{3} + x^{2} + 12 x^{2} + 12 x + 20 x + 20$ ... $[∵ 32 x = 12 x + 20 x]$

$= (x^{3} + x^{2}) + (12 x^{2} + 12 x) + (20 x + 20)$

$= x^{2} (x + 1) + 12 x (x + 1) + 20 (x + 1)$

$= (x + 1) (x^{2} + 12 x + 20)$

$= (x + 1) (x + 2) (x + 10)$ ... $[∵ x^{2} + 12 x + 20 = (x + 2) (x + 10)]$

Hence, $x^{3} + 13 x^{2} + 32 x + 20$ = $(x + 1) (x + 2) (x + 10)$ .

(iv) $2 y^{3} + y^{2} - 2 y - 1$

Answer

Let $p (y)$ = $2 y^{3} + y^{2} - 2 y - 1$

We shall now look for all the factors of $- 1$ .
These are $\pm 1$ .

Now, we find the values of $p (1), p (- 1)$ and we determine if any of these values are zero.

$∵ p (1) = 2 \times 1^{3} + 1^{2} - 2 \times 1 - 1$
$= 2 + 1 - 2 - 1$
$= 3 - 3$
$= 0$

$∴ (y - 1)$ is a factor of $p (y)$ .

Now, we divide $p (y)$ by $(y - 1)$ to get a quadratic polynomial as quotient.

So, $2 y^{3} + y^{2} - 2 y - 1$ $= (y - 1) (2 y^{2} + 3 y + 1)$
$= (y - 1) (2 y^{2} + 2 y + y + 1)$
$= (y - 1) [2 y (y + 1) + 1 (y + 1)]$
$= (y - 1) (y + 1) (2 y + 1)$

Or, we can also do the following to easily obtain the factors.

$2 y^{3} + y^{2} - 2 y - 1$

$= 2 y^{3} - 2 y^{2} + 3 y^{2} - 2 y - 1$ ... $[∵ y^{2} = - 2 y^{2} + 3 y^{2}]$

$= 2 y^{3} - 2 y^{2} + 3 y^{2} - 3 y + y - 1$ ... $[∵ y^{2} = - 2 y^{2} + 3 y^{2}]$

$= (2 y^{3} - 2 y^{2}) + (3 y^{2} - 3 y) + (y - 1)$

$= 2 y^{2} (y - 1) + 3 y (y - 1) + 1 (y - 1)$

$= (y - 1) (2 y^{2} + 3 y + 1)$

$= (y - 1) (2 y^{2} + 2 y + y + 1)$

$= (y - 1) [2 y (y + 1) + 1 (y + 1)]$

$= (y - 1) (y + 1) (2 y + 1)$ .

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Exercise 2.3 NCERT Class 9 Polynomials Mathematics Solutions

1. Determine which of the following polynomials has (x+1) a factor:

(i) x3+x2+x+1

Answer

(ii) x4+x3+x2+x+1

Answer

(iii) x4+3x3+3x2+x+1

Answer

(iv) x3−x2−(2+2)x+2

Answer

2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) p(x)=2x3+x2−2x−1, g(x)=x+1

Answer

(ii) p(x)=x3+3x2+3x+1, g(x)=x+2

Answer

(iii) p(x)=x3−4x2+x+6, g(x)=x−3

Answer

3. Find the value of k , if (x−1) is a factor of p(x) in each of the following cases:.

(i) p(x)=x2+x+k

Answer

(ii) p(x)=2x2+kx+2

Answer

(iii) p(x)=kx2−2x+1

Answer

(iv) p(x)=kx2−3x+k

Answer

4. Factorise:

(i) 12x2−7x+1

Answer

(ii) 2x2+7x+3

Answer

(iii) 6x2+5x−6

Answer

(iv) 3x2−x−4

Answer

5. Factorise:

(i) x3−2x2−x+2

Answer

(ii) x3−3x2−9x−5

Answer

(iii) x3+13x2+32x+20

Answer

(iv) 2y3+y2−2y−1

Answer

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1. Determine which of the following polynomials has $(x + 1)$ a factor:

(i) $x^{3} + x^{2} + x + 1$

(ii) $x^{4} + x^{3} + x^{2} + x + 1$

(iii) $x^{4} + 3 x^{3} + 3 x^{2} + x + 1$

(iv) $x^{3} - x^{2} - (2 + \sqrt{2}) x + \sqrt{2}$

2. Use the Factor Theorem to determine whether $g (x)$ is a factor of $p (x)$ in each of the following cases:

(i) $p (x) = 2 x^{3} + x^{2} - 2 x - 1,$ $g (x) = x + 1$

(ii) $p (x) = x^{3} + 3 x^{2} + 3 x + 1,$ $g (x) = x + 2$

(iii) $p (x) = x^{3} - 4 x^{2} + x + 6,$ $g (x) = x - 3$

3. Find the value of $k$ , if $(x - 1)$ is a factor of $p (x)$ in each of the following cases:.

(i) $p (x) = x^{2} + x + k$

(ii) $p (x) = 2 x^{2} + k x + \sqrt{2}$

(iii) $p (x) = k x^{2} - \sqrt{2} x + 1$

(iv) $p (x) = k x^{2} - 3 x + k$

(i) $12 x^{2} - 7 x + 1$

(ii) $2 x^{2} + 7 x + 3$

(iii) $6 x^{2} + 5 x - 6$

(iv) $3 x^{2} - x - 4$

(i) $x^{3} - 2 x^{2} - x + 2$

(ii) $x^{3} - 3 x^{2} - 9 x - 5$

(iii) $x^{3} + 13 x^{2} + 32 x + 20$

(iv) $2 y^{3} + y^{2} - 2 y - 1$