Exercise 2D RS Aggarwal Class 9 Polynomials Best Solutions

Exercise 2D RS Aggarwal Class 9 contains a total of twenty six questions. The questions are based on the following topic.

Factor Theorem: Let $p (x)$ be a polynomial of degree 1 or more and let $β$ be any real number.

If $p (β) = 0$ then $(x - β)$ is a factor of $p (x)$ .
If $(x - β)$ is a factor of $p (x)$ then $p (β) = 0$ .

Exercise 2D RS Aggarwal Class 9 Mathematics Solutions

Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7 Q.8 Q.9 Q.10 Q.11 Q.12 Q.13 Q.14 Q.15 Q.16 Q.17 Q.18 Q.19 Q.20 Q.21 Q.22 Q.23 Q.24 Q.25 Q.26

The questions from 1 to 10 in Exercise 2D RS Aggarwal Class 9 revolves around applications of factor theorem.

Using factor theorem, show that $g (x)$ is a factor of $p (x)$ , when

1. $p (x) = x^{3} - 8, g (x) = x - 2$

Answer

We have
$p (x) = x^{3} - 8$
$g (x) = x - 2$
Here, $g (x) = 0$
$\Rightarrow x - 2 = 0$
$\Rightarrow x = 2$

According to Factor Theorem, if $p (2) = 0$ , $g (x)$ is a factor of $p (x)$ .

$∵ p (2) = 2^{3} - 8 = 8 - 8 = 0$
$∴ g (x) = (x - 2)$ is a factor of $p (x)$ .

Hence, proved!

2. $p (x) = 2 x^{3} + 7 x^{2} - 24 x - 45, g (x) = x - 3$

Answer

We have
$p (x) = 2 x^{3} + 7 x^{2} - 24 x - 45$ ,
$g (x) = x - 3$
Here, $g (x) = 0$
$\Rightarrow x - 3 = 0$
$\Rightarrow x = 3$

According to Factor Theorem, if $p (3) = 0$ , $g (x)$ is a factor of $p (x)$ .

$∵ p (3)$
$= 2 \times 3^{3} + 7 \times 3^{2} - 24 \times 3 - 45$
$= 2 \times 27 + 7 \times 9 - 72 - 45$
$= 54 + 63 - 72 - 45$
$= 117 - 117$
$= 0$
$∴ g (x) = (x - 3)$ is a factor of $p (x)$ .

Hence, proved!

3. $p (x) = 2 x^{4} + 9 x^{3} + 6 x^{2} - 11 x - 6, g (x) = x - 1$

Answer

We have
$p (x) = 2 x^{4} + 9 x^{3} + 6 x^{2} - 11 x - 6$
$g (x) = x - 1$
Here, $g (x) = 0$
$\Rightarrow x - 1 = 0$
$\Rightarrow x = 1$

According to Factor Theorem, if $p (1) = 0$ , $g (x)$ is a factor of $p (x)$ .

$∵ p (1)$
$= 2 \times 1^{4} + 9 \times 1^{3} + 6 \times 1^{2} - 11 \times 1 - 6$
$= 2 + 9 + 6 - 11 - 6$
$= 15 - 15$
$= 0$
$∴ g (x) = (x - 1)$ is a factor of $p (x)$ .

Hence, proved!

4. $p (x) = x^{4} - x^{2} - 12, g (x) = x + 2$

Answer

We have
$p (x) = x^{4} - x^{2} - 12$
$g (x) = x + 2$
Here, $g (x) = 0$
$\Rightarrow x + 2 = 0$
$\Rightarrow x = - 2$

According to Factor Theorem, if $p (- 2) = 0$ , $g (x)$ is a factor of $p (x)$ .

$∵ p (- 2)$
$= {(- 2)}^{4} - {(- 2)}^{2} - 12$
$= 16 - 4 - 12$
$= 16 - 16$
$= 0$
$∴ g (x) = (x + 2)$ is a factor of $p (x)$ .

Hence, proved!

5. $p (x) = 69 + 11 x - x^{2} + x^{3}, g (x) = x + 3$

Answer

We have
$p (x) = 69 + 11 x - x^{2} + x^{3}$
$g (x) = x + 3$

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

According to Factor Theorem, if $p (- 3) = 0$ , $g (x)$ is a factor of $p (x)$ .

$∵ p (- 3)$
$= 69 + 11 \times (- 3) - {(- 3)}^{2} + {(- 3)}^{3}$
$= 69 - 33 - 9 - 27$
$= 69 - 69$
$= 0$
$∴ g (x) = (x + 3)$ is a factor of $p (x)$ .

Hence, proved!

6. $p (x) = 2 x^{3} + 9 x^{2} - 11 x - 30, g (x) = x + 5$

Answer

We have
$p (x) = 2 x^{3} + 9 x^{2} - 11 x - 30$
$g (x) = x + 5$

Here, $g (x) = 0$
$\Rightarrow x + 5 = 0$
$\Rightarrow x = - 5$

According to Factor Theorem, if $p (- 5) = 0$ , $g (x)$ is a factor of $p (x)$ .

$∵ p (- 5)$
$= 2 \times {(- 5)}^{3} + 9 \times {(- 5)}^{2} - 11 \times (- 5) - 30$
$= 2 \times (- 125) + 9 \times 25 + 55 - 30$
$= - 250 + 225 + 55 - 30$
$= - 280 + 280$
$= 0$
$∴ g (x) = (x + 5)$ is a factor of $p (x)$ .

Hence, proved!

7. $p (x) = 2 x^{4} + x^{3} - 8 x^{2} - x + 6, g (x) = 2 x - 3$

Answer

We have
$p (x) = 2 x^{4} + x^{3} - 8 x^{2} - x + 6$
$g (x) = 2 x - 3$

Here, $g (x) = 0$
$\Rightarrow 2 x - 3 = 0$
$\Rightarrow x = \frac{3}{2}$

According to Factor Theorem, if $p (\frac{3}{2}) = 0$ , $g (x)$ is a factor of $p (x)$ .

$∵ p (\frac{3}{2})$
$= 2 \times {(\frac{3}{2})}^{4} + {(\frac{3}{2})}^{3} - 8 \times {(\frac{3}{2})}^{2} - (\frac{3}{2}) + 6$
$= 2 \times \frac{81}{16} + \frac{27}{8} - 8 \times \frac{9}{4} - \frac{3}{2} + 6$
$= \frac{81}{8} + \frac{27}{8} - 18 - \frac{3}{2} + 6$
$= \frac{81 + 27 - 144 - 12 + 48}{8}$
$= \frac{156 - 156}{8}$
$= \frac{0}{8}$
$= 0$
$∴ g (x) = (2 x - 3)$ is a factor of $p (x)$ .

Hence, proved!

8. $p (x) = 3 x^{3} + x^{2} - 20 x + 12, g (x) = 3 x - 2$

Answer

We have
$p (x) = 3 x^{3} + x^{2} - 20 x + 12$
$g (x) = 3 x - 2$

Here, $g (x) = 0$
$\Rightarrow 3 x - 2 = 0$
$\Rightarrow x = \frac{2}{3}$

According to Factor Theorem, if $p (\frac{2}{3}) = 0$ , $g (x)$ is a factor of $p (x)$ .

$∵ p (\frac{2}{3})$
$= 3 \times {(\frac{2}{3})}^{3} + {(\frac{2}{3})}^{2} - 20 \times (\frac{2}{3}) + 12$
$= 3 \times \frac{8}{27} + \frac{4}{9} - \frac{40}{3} + 12$
$= \frac{8}{9} + \frac{4}{9} - \frac{40}{3} + 12$
$= \frac{8 + 4 - 120 + 108}{9}$
$= \frac{120 - 120}{9}$
$= \frac{0}{9} = 0$
$∴ g (x) = (3 x - 2)$ is a factor of $p (x)$ .

Hence, proved!

9. $p (x) = 7 x^{2} - 4 \sqrt{2} x - 6, g (x) = x - \sqrt{2}$

Answer

We have
$p (x) = 7 x^{2} - 4 \sqrt{2} x - 6$
$g (x) = x - \sqrt{2}$

Here, $g (x) = 0$
$\Rightarrow x - \sqrt{2} = 0$
$\Rightarrow x = \sqrt{2}$ .

According to Factor Theorem, if $p (\sqrt{2}) = 0$ , $g (x)$ is a factor of $p (x)$ .

$∵ p (\sqrt{2})$
$= 7 \times {(\sqrt{2})}^{2} - 4 \times \sqrt{2} \times (\sqrt{2}) - 6$
$= 7 \times 2 - 4 \times 2 - 6$
$= 14 - 8 - 6$
$= 14 - 14$
$= 0$
$∴ g (x) = (x - \sqrt{2})$ is a factor of $p (x)$ .

Hence, proved!

10. $p (x) = 2 \sqrt{2} x^{2} + 5 x + \sqrt{2}, g (x) = x + \sqrt{2}$

Answer

We have
$p (x) = 2 \sqrt{2} x^{2} + 5 x + \sqrt{2}$
$g (x) = x + \sqrt{2}$

$∵ p (x)$ is exactly divisible by $(x + 2)$ .
$∴ (x + 2)$ is a factor of $p (x)$
$\Rightarrow p (- 2) = 0$
$\Rightarrow {(- 2)}^{4} + a {(- 2)}^{3} - 7 {(- 2)}^{2} - 8 \times (- 2) + b = 0$
$\Rightarrow 16 - 8 a - 7 \times 4 + 16 + b = 0$
$\Rightarrow 16 - 8 a - 28 + 16 + b = 0$
$\Rightarrow - 8 a + b + 4 = 0$
$\Rightarrow - 8 a + b = - 4$
$\Rightarrow 8 a - b = 4$ ... (1)

Here, $g (x) = 0$
$\Rightarrow x + \sqrt{2} = 0$
$\Rightarrow x = - \sqrt{2}$ .

According to Factor Theorem, if $p (- \sqrt{2}) = 0$ , $g (x)$ is a factor of $p (x)$ .

$∵ p (- \sqrt{2})$
$= 2 \sqrt{2} \times {(- \sqrt{2})}^{2} + 5 \times (- \sqrt{2}) + \sqrt{2}$
$= 4 \sqrt{2} - 5 \sqrt{2} + \sqrt{2}$
$= - \sqrt{2} + \sqrt{2}$
$= 0$
$∴ g (x) = (x + \sqrt{2})$ is a factor of $p (x)$ .

Hence, proved!

The 11th question of Exercise 2D RS Aggarwal Class 9 asks us to show if a polynomial is a factor of another polynomial.

11. Show that $(p - 1)$ is a factor of $(p^{10} - 1)$ and also of $(p^{11} - 1)$

Answer

Let
$f (p) = p^{10} - 1$
$g (p) = p^{11} - 1$
$h (p) = p - 1$

Here, $h (p) = 0$
$\Rightarrow p - 1 = 0$
$\Rightarrow p = 1$

Now,
$f (1) = 1^{10} - 1 = 1 - 1 = 0$
$\Rightarrow p - 1$ is a factor of $f (p)$ . --> [Factor Theorem]
$g (1) = 1^{11} - 1 = 1 - 1 = 0$
$\Rightarrow p - 1$ is a factor of $g (p)$ . -->[Factor Theorem]

Hence, proved

In Exercise 2D RS Aggarwal Class 9, the twelfth question asks us to figure out the value of a variable when a factor of a polynomial is given.

12. Find the value of $k$ for which $(x - 1)$ is a factor of $2 x^{3} + 9 x^{2} + x + k$ .

Answer

Let
$p (x) = 2 x^{3} + 9 x^{2} + x + k$
$g (x) = x - 1$
Here, $g (x) = 0$
$\Rightarrow x - 1 = 0$
$\Rightarrow x = 1$

According to factor theorem

If $(x - a)$ is a factor of $f (x)$ , then $f (a) = 0$ .

$∵ g (x) = (x - 1)$ is a factor of $p (x)$ .
$∴ p (1) = 0$
$\Rightarrow 2 \times 1^{3} + 9 \times 1^{2} + 1 + k = 0$
$\Rightarrow 2 \times 1 + 9 \times 1 + 1 + k = 0$
$\Rightarrow 2 + 9 + 1 + k = 0$
$\Rightarrow 12 + k = 0$
$\Rightarrow k = - 12$

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

In Exercise 2D RS Aggarwal Class 9, the thirteenth question asks us to figure out the value of a variable when a factor of a polynomial is given.

13. Find the value of $a$ for which $(x - 4)$ is a factor of $2 x^{3} - 3 x^{2} - 18 x + a$ .

Answer

Let
$p (x) = 2 x^{3} - 3 x^{2} - 18 x + a$
$g (x) = x - 4$
Here, $g (x) = 0$
$\Rightarrow x - 4 = 0$
$\Rightarrow x = 4$

According to factor theorem

If $(x - a)$ is a factor of $f (x)$ , then $f (a) = 0$ .

$∵ g (x) = (x - 4)$ is a factor of $p (x)$ .
$∴ p (4) = 0$
$\Rightarrow 2 \times 4^{3} - 3 \times 4^{2} - 18 \times 4 + a = 0$
$\Rightarrow 2 \times 64 - 3 \times 16 - 72 + a = 0$
$\Rightarrow 128 - 48 - 72 + a = 0$
$\Rightarrow 128 - 120 + a = 0$
$\Rightarrow 8 + a = 0$
$\Rightarrow a = - 8$

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

In Exercise 2D RS Aggarwal Class 9, the fourteenth question asks us to find the value of a variable if a factor of a polynomial is given.

14. Find the value of $a$ for which $(x + 1)$ is a factor of $(a x^{3} + x^{2} - 2 x + 4 a - 9)$ .

Answer

Let
$p (x) = a x^{3} + x^{2} - 2 x + 4 a - 9$
$g (x) = x + 1$
Here, $g (x) = 0$
$\Rightarrow x + 1 = 0$
$\Rightarrow x = - 1$ .

According to factor theorem

If $(x - b)$ is a factor of $f (x)$ , then $f (b) = 0$ .

$∵ g (x) = (x + 1)$ is a factor of $p (x)$ .
$∴ p (- 1) = 0$
$\Rightarrow a {(- 1)}^{3} + {(- 1)}^{2} - 2 (- 1) + 4 a - 9 = 0$
$\Rightarrow - a + 1 + 2 + 4 a - 9 = 0$
$\Rightarrow 3 a - 6 = 0$
$\Rightarrow 3 a = 6$
$\Rightarrow a = \frac{6}{3}$
$\Rightarrow a = 2$

Hence, the value of $a$ is $2$ .

In Exercise 2D RS Aggarwal Class 9, the fifteenth question asks us to find the value of a variable when a factor of a polynomial is known.

15. Find the value of $a$ for which $(x + 2 a)$ is a factor of $(x^{5} - 4 a^{2} x^{3} + 2 x + 2 a + 3)$ .

Answer

Let
$p (x) = x^{5} - 4 a^{2} x^{3} + 2 x + 2 a + 3$
$g (x) = x + 2 a$
Here, $g (x) = 0$
$\Rightarrow x + 2 a = 0$
$\Rightarrow x = - 2 a$ .

According to factor theorem

If $(x - b)$ is a factor of $f (x)$ , then $f (b) = 0$ .

$∵ g (x) = (x + 2 a)$ is a factor of $p (x)$ .
$∴ p (- 2 a) = 0$
$\Rightarrow {(- 2 a)}^{5} - 4 a^{2} {(- 2 a)}^{3} + 2 (- 2 a) + 2 a + 3 = 0$
$\Rightarrow {(- 2)}^{5} \times a^{5} - 4 a^{2} \times {(- 2)}^{3} \times a^{3} + 2 (- 2 a) + 2 a + 3 = 0$
$\Rightarrow - 32 a^{5} - 4 a^{2} \times (- 8) \times a^{3} - 4 a + 2 a + 3 = 0$
$\Rightarrow - 32 a^{5} + 32 a^{5} - 2 a + 3 = 0$
$\Rightarrow - 2 a + 3 = 0$
$\Rightarrow - 2 a = - 3$
$\Rightarrow a = \frac{3}{2}$

Hence, the value of $a$ is $\frac{3}{2}$ .

In Exercise 2D RS Aggarwal Class 9, the sixteenth question asks us to find the value of a variable when a factor of a polynomial is known.

16. Find the value of $m$ for which $(2 x - 1)$ is a factor of $(8 x^{4} + 4 x^{3} - 16 x^{2} + 10 x + m)$ .

Answer

Let
$p (x) = 8 x^{4} + 4 x^{3} - 16 x^{2} + 10 x + m$
$g (x) = (2 x - 1)$
Here, $g (x) = 0$
$\Rightarrow (2 x - 1) = 0$
$\Rightarrow x = \frac{1}{2}$ .

According to factor theorem

If $(x - b)$ is a factor of $f (x)$ , then $f (b) = 0$ .

$∵ g (x) = (2 x - 1)$ is a factor of $p (x)$ .
$∴ p (\frac{1}{2}) = 0$
$\Rightarrow 8 {(\frac{1}{2})}^{4} + 4 {(\frac{1}{2})}^{3} - 16 {(\frac{1}{2})}^{2} + 10 (\frac{1}{2}) + m = 0$
$\Rightarrow 8 \times \frac{1}{16} + 4 \times \frac{1}{8} - 16 \times \frac{1}{4} + 10 \times \frac{1}{2} + m = 0$
$\Rightarrow \frac{1}{2} + \frac{1}{2} - 4 + 5 + m = 0$
$\Rightarrow 1 - 4 + 5 + m = 0$
$\Rightarrow 2 + m = 0$
$\Rightarrow m = - 2$

Hence, the value of $m$ is $- 2$ .

In Exercise 2D RS Aggarwal Class 9, the seventeenth question asks us to find the value of a variable.

17. Find the value of $a$ for which the polynomial $(x^{4} - x^{3} - 11 x^{2} - x + a)$ is divisible by $(x + 3)$ .

Answer

Let
$p (x) = x^{4} - x^{3} - 11 x^{2} - x + a$
$g (x) = (x + 3)$
Here, $g (x) = 0$
$\Rightarrow (x + 3) = 0$
$\Rightarrow x = - 3$ .

According to factor theorem

If $f (x)$ is divisible by $(x - a)$ , then $f (a) = 0$ .

$∵ p (x)$ is divisible by $(x + 3)$ .
It means $(x + 3)$ is a factor of $p (x)$ .
$∴ p (- 3) = 0$
$\Rightarrow {(- 3)}^{4} - {(- 3)}^{3} - 11 {(- 3)}^{2} - (- 3) + a = 0$
$\Rightarrow 81 + 27 - 11 \times 9 + 3 + a = 0$
$\Rightarrow 81 + 27 - 99 + 3 + a = 0$
$\Rightarrow 12 + a = 0$
$\Rightarrow a = - 12$

Hence, the value of $a$ is $- 12$ .

In Exercise 2D RS Aggarwal Class 9, the eighteenth question asks us to show that a given polynomial is a divisor of the given polynomial.

18. Without actual division, show that $(x^{3} - 3 x^{2} - 13 x + 15)$ is exactly divisible by $(x^{2} + 2 x - 3)$ .

Answer

Let
$p (x) = x^{3} - 3 x^{2} - 13 x + 15$
$g (x) = (x^{2} + 2 x - 3)$
Here, $g (x) = 0$
$\Rightarrow x^{2} + 2 x - 3 = 0$
$\Rightarrow x^{2} + 3 x - x - 3 = 0$
$\Rightarrow x (x + 3) - 1 (x + 3) = 0$
$\Rightarrow (x + 3) (x - 1) = 0$
$\Rightarrow x + 3 = 0 o r x - 1 = 0$
$\Rightarrow x = - 3 o r x = 1$

Putting $x = - 3$ in $p (x)$ , we get.
$∵ p (x) = x^{3} - 3 x^{2} - 13 x + 15$ $\Rightarrow p (- 3) = {(- 3)}^{3} - 3 \times {(- 3)}^{2} - 13 \times (- 3) + 15$
$\Rightarrow p (- 3) = - 27 - 3 \times 9 + 39 + 15$
$\Rightarrow p (- 3) = - 54 + 54$
$\Rightarrow p (- 3) = 0$

Putting $x = 1$ in $p (x)$ , we get.
$∵ p (x) = x^{3} - 3 x^{2} - 13 x + 15$ $\Rightarrow p (1) = {(1)}^{3} - 3 \times {(1)}^{2} - 13 \times (1) + 15$
$\Rightarrow p (1) = 1 - 3 - 13 + 15$
$\Rightarrow p (1) = 16 - 16$
$\Rightarrow p (1) = 0$

$∵ p (- 3) = 0, p (1) = 0$ .
$∴ (x + 3) a n d (x - 1)$ divides $p (x)$ exactly.
$\Rightarrow (x + 3) (x - 1)$ divides $p (x)$ .
$\Rightarrow x^{2} + 2 x - 3$ divides $p (x)$
In other words,
$p (x)$ is exactly divisible by ( $x^{2} + 2 x - 3$ ).
Hence, proved!

In Exercise 2D RS Aggarwal Class 9, the ninth question asks us to find the value of two variables at the given conditions.

19. If $(x^{3} + a x^{2} + b x + 6)$ 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 $p (x) = x^{3} + a x^{2} + b x + 6$ .

$∵ (x - 2)$ is a factor of $p (x)$ ... (Given)
$∴ 2$ is a zero of the polynomial $p (x)$ .
$\Rightarrow p (2) = 0$
$\Rightarrow 2^{3} + a \times 2^{2} + b \times 2 + 6 = 0$
$\Rightarrow 8 + 4 a + 2 b + 6 = 0$
$\Rightarrow 4 a + 2 b + 14 = 0$
$\Rightarrow 2 a + b + 7 = 0$
$\Rightarrow 2 a + b = - 7$ ... (1)

It is given that $p (x)$ leaves remainder 3 when divided by $(x - 3)$ .
$\Rightarrow p (3) = 3$ ... (Remainder Theorem)
$\Rightarrow 3^{3} + a \times 3^{2} + b \times 3 + 6 = 3$
$\Rightarrow 27 + 9 a + 3 b + 6 = 3$
$\Rightarrow 9 a + 3 b + 33 = 3$
$\Rightarrow 9 a + 3 b = 3 - 33$
$\Rightarrow 9 a + 3 b = - 30$
$\Rightarrow 3 a + b = - 10$ ... (2)

Solving equations (1) and (2) by elimination method, we get.

2 a + b = - 7

3 a + b = - 10

− − +

- a = 3

\Rightarrow a = - 3

Putting the value of $a$ in equation (1), we get.

$2 \times (- 3) + b = - 7$
$\Rightarrow - 6 + b = - 7$
$\Rightarrow b = - 7 + 6$
$\Rightarrow b = - 1$

Hence, $a = - 3, b = - 1$ .

In Exercise 2D RS Aggarwal Class 9, the twentieth question is about finding values of variables when a polynomial is exactly divisible by $(x - 1)$ as well as $(x - 2)$ .

20. Find the values of $a$ and $b$ so that the polynomial $(x^{3} - 10 x^{2} + a x + b)$ is exactly divisible by $(x - 1)$ as well as $(x - 2)$ .

Answer

Let $p (x) = x^{3} - 10 x^{2} + a x + b$

$∵ p (x)$ is exactly divisible by $(x - 1)$ .
$∴ (x - 1)$ is a factor of $p (x)$
$\Rightarrow p (1) = 0$
$\Rightarrow 1^{3} - 10 \times 1^{2} + a \times 1 + b = 0$
$\Rightarrow 1 - 10 + a + b = 0$
$\Rightarrow - 9 + a + b = 0$
$\Rightarrow a + b = 9$ ... (1)

$∵ p (x)$ is exactly divisible by $(x - 2)$ .
$∴ (x - 2)$ is a factor of $p (x)$
$\Rightarrow p (2) = 0$
$\Rightarrow 2^{3} - 10 \times 2^{2} + a \times 2 + b = 0$
$\Rightarrow 8 - 10 \times 4 + 2 a + b = 0$
$\Rightarrow 8 - 40 + 2 a + b = 0$
$\Rightarrow - 32 + 2 a + b = 0$
$\Rightarrow 2 a + b = 32$ ... (2)

Solving equations (1) and (2) by elimination method, we get.

a + b = 9

2 a + b = 32

− − −

- a = - 23

\Rightarrow a = 23

Putting the value of $a$ in equation (1), we get.

$23 + b = 9$
$\Rightarrow b = 9 - 23$
$\Rightarrow b = - 14$

Hence, $a = 23, b = - 14$ .

In Exercise 2D RS Aggarwal Class 9, the 21st question asks us to find the value of two variable at a given condition.

21. Find the values of $a$ and $b$ so that the polynomial $(x^{4} + a x^{3} - 7 x^{2} - 8 x + b)$ is exactly divisible by $(x + 2)$ as well as $(x + 3)$ .

Answer

Let $p (x) = (x^{4} + a x^{3} - 7 x^{2} - 8 x + b)$

$∵ p (x)$ is exactly divisible by $(x + 3)$ .
$∴ (x + 3)$ is a factor of $p (x)$
$\Rightarrow p (- 3) = 0$
$\Rightarrow {(- 3)}^{4} + a {(- 3)}^{3} - 7 {(- 3)}^{2} - 8 \times (- 3) + b = 0$
$\Rightarrow 81 - 27 a - 7 \times 9 + 24 + b = 0$
$\Rightarrow 81 - 27 a - 63 + 24 + b = 0$
$\Rightarrow - 27 a + b + 42 = 0$
$\Rightarrow - 27 a + b = - 42$
$\Rightarrow 27 a - b = 42$ ... (2)

Solving equations (1) and (2) by elimination method, we get.

8 a - b = 4

27 a - b = 42

- + -

- 19 a = - 38

\Rightarrow a = \frac{- 38}{- 19}

\Rightarrow a = 2

Putting the value of $a$ in equation (1), we get.

$8 \times 2 - b = 4$
$\Rightarrow 16 - b = 4$
$\Rightarrow - b = 4 - 16$
$\Rightarrow - b = - 12$
$\Rightarrow b = 12$

Hence, $a = 2, b = 12$ .

In Exercise 2D RS Aggarwal Class 9, the 22nd question asks us to prove an equality at a given condition.

22. If both $(x - 2)$ and $(x - \frac{1}{2})$ are factors of $p x^{2} + 5 x + r$ , prove that $p = r$ .

Answer

Let $m (x) = p x^{2} + 5 x + r$ .

$∵ (x - 2) & (x - \frac{1}{2})$ are factors of $m (x)$ .
$\Rightarrow x = 2 a n d x = \frac{1}{2}$ are zeros of $m (x)$ .
$\Rightarrow m (2) = 0 & m (\frac{1}{2}) = 0$
$\Rightarrow p \times 2^{2} + 5 \times 2 + r = 0 & p \times {(\frac{1}{2})}^{2} + 5 \times \frac{1}{2} + r = 0$
$\Rightarrow 4 p + 10 + r = 0 & \frac{p}{4} + \frac{5}{2} + r = 0$
$\Rightarrow 4 p + 10 + r = 0 & \frac{p + 10 + 4 r}{4} = 0$
$\Rightarrow 4 p + 10 + r = 0 & p + 10 + 4 r = 0$
From both the equations, we can see that both the expressions on the left hand sides of the equations are equal to 0.
$\Rightarrow 4 p + 10 + r = p + 10 + 4 r$
$\Rightarrow 4 p - p + 10 - 10 = 4 r - r$
$\Rightarrow 3 p = 3 r$
$\Rightarrow p = r$

Hence, proved!

In Exercise 2D RS Aggarwal Class 9, the 23rd question asks us to prove that a polynomial is divisible by another polynomial.

23. Without actual division, prove that $(2 x^{4} - 5 x^{3} + 2 x^{2} - x + 2)$ is divisible by $(x^{2} - 3 x + 2)$ .

Answer

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

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

Clearly, $p (x)$ will be divisible by $g (x)$ only when it is exactly divisible by $(x - 1)$ as well as $(x - 2)$ .

Now, $(x - 1) = 0$ $\Rightarrow x = 1$
and $(x - 2) = 0$ $\Rightarrow x = 2$

By factor theorem, $g (x)$ will be a factor of $p (x)$ , if $p (1) = 0$ and $p (2) = 0$ .

Now, $p (1) = 0$
$\Rightarrow p (1) = 2 \times 1^{4} - 5 \times 1^{3} + 2 \times 1^{2} - 1 + 2$
$= 2 - 5 + 2 - 1 + 2$
$= 6 - 6$
$= 0$

And, $p (2) = 0$
$\Rightarrow p (2) = 2 \times 2^{4} - 5 \times 2^{3} + 2 \times 2^{2} - 2 + 2$
$= 2 \times 16 - 5 \times 8 + 2 \times 4 - 2 + 2$
$= 32 - 40 + 8 - 2 + 2$
$= 42 - 42$
$= 0$

Thus, $p (x)$ is exactly divisible by each one of $(x - 1)$ and $(x - 2)$ .

Hence, $p (x)$ is exactly divisible by $(x - 1) (x - 2)$ , i.e., by $x^{2} - 3 x + 2$ .

Hence, proved!

In Exercise 2D RS Aggarwal Class 9, the 24th question is a bit tricky. It asks us to add a constant to a polynomial so that it becomes divisible by another polynomial.

24. What must be added to $(2 x^{4} - 5 x^{3} + 2 x^{2} - x - 3)$ so that the result is exactly divisible by $(x - 2)$ .

Answer

When the given polynomial is divided by a linear polynomial then the remainder is constant.

Let the required number to be added be $k$ .
Let $p (x) = 2 x^{4} - 5 x^{3} + 2 x^{2} - x - 3 + k$ and
$g (x) = (x - 2)$ .

Then, $g (x) = 0$
$\Rightarrow x - 2 = 0$
$\Rightarrow x = 2$ .

By factor theorem, $p (x)$ will be divisible by $(x - 2)$ , if $p (2) = 0$ .

Now, $p (2) = 0$
$\Rightarrow 2 \times 2^{4} - 5 \times 2^{3} + 2 \times 2^{2} - 2 - 3 + k = 0$
$\Rightarrow 2 \times 16 - 5 \times 8 + 2 \times 4 - 2 - 3 + k = 0$
$\Rightarrow 32 - 40 + 8 - 2 - 3 + k = 0$
$\Rightarrow - 5 + k = 0$
$\Rightarrow k = 5$

Hence, the required number to be added is 5.

In Exercise 2D RS Aggarwal Class 9, the 25th question asks us to find the value of constant which must be added to the polynomial so that it becomes divisible by another polynomial.

25. What must be subtracted from $(x^{4} + 2 x^{3} - 2 x^{2} + 4 x + 6)$ so that the result is exactly divisible by $(x^{2} + 2 x - 3)$ ?

Answer

When the given polynomial is divided by a quadratic polynomial, then the remainder is a linear expression. So, we need to subtract a linear expression to get the result which is exactly divisible by the given quadratic polynomial.

Let that linear expression that is to be subtracted is $a x + b$ .

Let $p (x) = (x^{4} + 2 x^{3} - 2 x^{2} + 4 x + 6) - (a x + b)$
$= x^{4} + 2 x^{3} - 2 x^{2} + 4 x + 6 - a x - b$
$= x^{4} + 2 x^{3} - 2 x^{2} + 4 x - a x - b + 6$
$= x^{4} + 2 x^{3} - 2 x^{2} + x (4 - a) - b + 6$
and $g (x) = (x^{2} + 2 x - 3)$

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

Now, $p (x)$ will be divisible by $g (x)$ only when it is divisible by $(x + 3)$ as well as $(x - 1)$ .

Now, $(x + 3) = 0$
$\Rightarrow x = - 3$
and $(x - 1) = 0$
$\Rightarrow x = 1$

By factor theorem, $p (x)$ will be divisible by $g (x)$ , if $p (- 3) = 0$ and $p (1) = 0$ .

$p (- 3) = 0$
$\Rightarrow {(- 3)}^{4} + 2 \times {(- 3)}^{3} - 2 \times {(- 3)}^{2} + (- 3) (4 - a) - b + 6 = 0$
$\Rightarrow 81 - 54 - 18 - 12 + 3 a - b + 6 = 0$
$\Rightarrow 3 a - b + 3 = 0$
$\Rightarrow 3 a - b = - 3$ ... (1)

$p (1) = 0$
$\Rightarrow 1^{4} + 2 \times 1^{3} - 2 \times 1^{2} + 1 \times (4 - a) - b + 6 = 0$
$\Rightarrow 1 + 2 - 2 + 4 - a - b + 6 = 0$
$\Rightarrow - a - b + 11 = 0$
$\Rightarrow a + b = 11$ ... (2)

Solving equations (1) and (2) by elimination method, we get.

3 a - b = - 3

a + b = 11

4 a = 8

\Rightarrow a = \frac{8}{4}

\Rightarrow a = 2

Putting the value of $a$ in equation (2), we get.
$2 + b = 11$
$\Rightarrow b = 11 - 2$
$\Rightarrow b = 9$

Hence, the required expression that is to be subtracted from the given polynomial is $2 x + 9$ .

In Exercise 2D RS Aggarwal Class 9, the 26th question asks us to prove a linear polynomial to be a factor of another polynomial.

26. Use factor theorem to prove that $(x + a)$ is a factor of $(x^{n} + a^{n})$ for any odd positive integer n.

Answer

Let $p (x) = (x^{n} + a^{n})$ , where n is any positive odd integer i.e. $n = 1, 3, 5, . . .$
and $g (x) = x + a$ .

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

By factor theorem, $(x + a)$ will be a factor of $p (x)$ , if $p (- a) = 0$ .

$p (- a) = [{(- a)}^{n} + a^{n}]$
$= [{(- 1)}^{n} \times a^{n} + a^{n}]$
$= [{(- 1)}^{n} + 1] a^{n}$
$= [- 1 + 1] \times a^{n}$ [ $∵ n$ being odd, ${(- 1)}^{n} = - 1$ ]
$= 0 \times a^{n}$
$= 0$

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

Hence, proved!

Exercise 2D RS Aggarwal Class 9 Mathematics Solutions

Using factor theorem, show that g(x) is a factor of p(x), when

1. p(x)=x3−8, g(x)=x−2

Answer

2. p(x)=2x3+7x2−24x−45, g(x)=x−3

Answer

3. p(x)=2x4+9x3+6x2−11x−6, g(x)=x−1

Answer

4. p(x)=x4−x2−12, g(x)=x+2

Answer

5. p(x)=69+11x−x2+x3, g(x)=x+3

Answer

6. p(x)=2x3+9x2−11x−30, g(x)=x+5

Answer

7. p(x)=2x4+x3−8x2−x+6, g(x)=2x−3

Answer

8. p(x)=3x3+x2−20x+12, g(x)=3x−2

Answer

9. p(x)=7x2−42x−6, g(x)=x−2

Answer

10. p(x)=22x2+5x+2, g(x)=x+2

Answer

11. Show that (p−1) is a factor of (p10−1) and also of (p11−1)

Answer

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

Answer

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

Answer

14. Find the value of a for which (x+1) is a factor of (ax3+x2−2x+4a−9).

Answer

15. Find the value of a for which (x+2a) is a factor of (x5−4a2x3+2x+2a+3).

Answer

16. Find the value of m for which (2x−1) is a factor of (8x4+4x3−16x2+10x+m).

Answer

17. Find the value of a for which the polynomial (x4−x3−11x2−x+a) is divisible by (x+3).

Answer

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

Answer

19. 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

20. 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

21. 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

22. If both (x−2) and (x−12) are factors of px2+5x+r, prove that p=r.

Answer

23. Without actual division, prove that (2x4−5x3+2x2−x+2) is divisible by (x2−3x+2).

Answer

24. What must be added to (2x4−5x3+2x2−x−3) so that the result is exactly divisible by (x−2).

Answer

25. What must be subtracted from (x4+2x3−2x2+4x+6) so that the result is exactly divisible by (x2+2x−3)?

Answer

26. Use factor theorem to prove that (x+a) is a factor of (xn+an) for any odd positive integer n.

Answer

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Using factor theorem, show that $g (x)$ is a factor of $p (x)$ , when

1. $p (x) = x^{3} - 8, g (x) = x - 2$

2. $p (x) = 2 x^{3} + 7 x^{2} - 24 x - 45, g (x) = x - 3$

3. $p (x) = 2 x^{4} + 9 x^{3} + 6 x^{2} - 11 x - 6, g (x) = x - 1$

4. $p (x) = x^{4} - x^{2} - 12, g (x) = x + 2$

5. $p (x) = 69 + 11 x - x^{2} + x^{3}, g (x) = x + 3$

6. $p (x) = 2 x^{3} + 9 x^{2} - 11 x - 30, g (x) = x + 5$

7. $p (x) = 2 x^{4} + x^{3} - 8 x^{2} - x + 6, g (x) = 2 x - 3$

8. $p (x) = 3 x^{3} + x^{2} - 20 x + 12, g (x) = 3 x - 2$

9. $p (x) = 7 x^{2} - 4 \sqrt{2} x - 6, g (x) = x - \sqrt{2}$

10. $p (x) = 2 \sqrt{2} x^{2} + 5 x + \sqrt{2}, g (x) = x + \sqrt{2}$

11. Show that $(p - 1)$ is a factor of $(p^{10} - 1)$ and also of $(p^{11} - 1)$

12. Find the value of $k$ for which $(x - 1)$ is a factor of $2 x^{3} + 9 x^{2} + x + k$ .

13. Find the value of $a$ for which $(x - 4)$ is a factor of $2 x^{3} - 3 x^{2} - 18 x + a$ .

14. Find the value of $a$ for which $(x + 1)$ is a factor of $(a x^{3} + x^{2} - 2 x + 4 a - 9)$ .

15. Find the value of $a$ for which $(x + 2 a)$ is a factor of $(x^{5} - 4 a^{2} x^{3} + 2 x + 2 a + 3)$ .

16. Find the value of $m$ for which $(2 x - 1)$ is a factor of $(8 x^{4} + 4 x^{3} - 16 x^{2} + 10 x + m)$ .

17. Find the value of $a$ for which the polynomial $(x^{4} - x^{3} - 11 x^{2} - x + a)$ is divisible by $(x + 3)$ .

18. Without actual division, show that $(x^{3} - 3 x^{2} - 13 x + 15)$ is exactly divisible by $(x^{2} + 2 x - 3)$ .

19. If $(x^{3} + a x^{2} + b x + 6)$ has $(x - 2)$ as a factor and leaves a remainder 3 when divided by $(x - 3)$ , find the values of $a$ and $b$ .

20. Find the values of $a$ and $b$ so that the polynomial $(x^{3} - 10 x^{2} + a x + b)$ is exactly divisible by $(x - 1)$ as well as $(x - 2)$ .

21. Find the values of $a$ and $b$ so that the polynomial $(x^{4} + a x^{3} - 7 x^{2} - 8 x + b)$ is exactly divisible by $(x + 2)$ as well as $(x + 3)$ .

22. If both $(x - 2)$ and $(x - \frac{1}{2})$ are factors of $p x^{2} + 5 x + r$ , prove that $p = r$ .

23. Without actual division, prove that $(2 x^{4} - 5 x^{3} + 2 x^{2} - x + 2)$ is divisible by $(x^{2} - 3 x + 2)$ .

24. What must be added to $(2 x^{4} - 5 x^{3} + 2 x^{2} - x - 3)$ so that the result is exactly divisible by $(x - 2)$ .

25. What must be subtracted from $(x^{4} + 2 x^{3} - 2 x^{2} + 4 x + 6)$ so that the result is exactly divisible by $(x^{2} + 2 x - 3)$ ?

26. Use factor theorem to prove that $(x + a)$ is a factor of $(x^{n} + a^{n})$ for any odd positive integer n.