Exercise 2C RS Aggarwal Class 9 Polynomial Best Solutions

Exercise 2C RS Aggarwal Class 9 contains a total of sixteen questions. The questions are based on the following topics.

Division Algorithm in Polynomials
Remainder Theorem

Exercise 2C 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

The first question in Exercise 2C RS Aggarwal Class 9 asks us to find the quotient and remainder by doing actual division.

1. By actual division, find the quotient and the remainder when $(x^{4} + 1)$ is divided by $(x - 1)$ .
Verify that remainder = $f (1)$ .

Answer

From the above division, we see that remainder when $(x^{4} + 1)$ is divided by $(x + 1)$ is 2 and quotient is $x^{3} + x^{2} + x + 1$ .

Verification:
$∵ f (x) = x^{4} + 1$
Putting $x = 1$ , we get.
$f (1) = 1^{4} + 1$ $= 1 + 1 = 2$ .
Hence, verified!

The second question of Exercise 2C RS Aggarwal Class 9 asks us to verify the division algorithm for the polynomials.

2. Verify the division algorithm for the polynomials
$p (x) = 2 x^{4} - 6 x^{3} + 2 x^{2} - x + 2$ and $g (x) = x + 2$ .

Answer

Given polynomials are
$p (x) = 2 x^{4} - 6 x^{3} + 2 x^{2} - x + 2$ and $g (x) = x + 2$ .

After dividing $p (x)$ by $g (x)$ , we get.
Quotient $q (x)$ = $2 x^{3} - 10 x^{2} + 22 x - 45$
Remainder $r (x)$ = 92

Verification of Division Algorithm:

To verify: $D i v i d e n d = D i v i s o r \times Q u o t i e n t + R e m a i n d e r$
i.e. $p (x) = g (x) \times q (x) + r (x)$

Taking Right Hand Side (RHS), we have,
$D i v i s o r \times Q u o t i e n t + R e m a i n d e r$
$= (x + 2) \times (2 x^{3} - 10 x^{2} + 22 x - 45) + 92$
$= [x (2 x^{3} - 10 x^{2} + 22 x - 45) + 2 (2 x^{3} - 10 x^{2} + 22 x - 45)] + 92$
$= 2 x^{4} - 10 x^{3} + 22 x^{2} - 45 x + 4 x^{3} - 20 x^{2} + 44 x - 90 + 92$
$= 2 x^{4} - 6 x^{3} + 2 x^{2} - x + 2$
= $p (x)$ Dividend.
Hence, verified!

The questions from third to twelfth from Exercise 2C RS Aggarwal Class 9 asks us to use remainder theorem to find remainder.

Using the remainder theorem, find the remainder, when $p (x)$ is divided by $g (x)$ , where

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

Answer

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

According to remainder theorem, when $p (x)$ is divided by $g (x)$ , the remainder is $p (1)$ .

$∵ p (x) = x^{3} - 6 x^{2} + 9 x + 3$
$\Rightarrow p (1) = 1^{3} - 6 \times 1^{2} + 9 \times 1 + 3$
$\Rightarrow p (1) = 1 - 6 + 9 + 3$
$\Rightarrow p (1) = 7$

Hence, the remainder when $p (x)$ is divided by $g (x)$ is 7.

4. $p (x) = 2 x^{3} - 7 x^{2} + 9 x - 13, g (x) = x - 3$

Answer

We have
$p (x) = 2 x^{3} - 7 x^{2} + 9 x - 13, g (x) = x - 3$
Here, $g (x) = 0$
$\Rightarrow x - 3 = 0$
$\Rightarrow x = 3$

According to remainder theorem, when $p (x)$ is divided by $g (x)$ , the remainder is $p (3)$ .

$∵ p (x) = 2 x^{3} - 7 x^{2} + 9 x - 13$
$\Rightarrow p (3) = 2 \times 3^{3} - 7 \times 3^{2} + 9 \times 3 - 13$
$\Rightarrow p (3) = 54 - 63 + 27 - 13$
$\Rightarrow p (3) = 5$ .

Hence, the remainder when $p (x)$ is divided by $g (x)$ is 5.

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

Answer

We have
$p (x) = 3 x^{4} - 6 x^{2} - 8 x - 2, g (x) = x - 2$
Here, $g (x) = 0$
$\Rightarrow x - 2 = 0$
$\Rightarrow x = 2$ .

According to remainder theorem, when $p (x)$ is divided by $g (x)$ , the remainder is $p (2)$ .

$∵ p (x) = 3 x^{4} - 6 x^{2} - 8 x - 2$
$\Rightarrow p (2) = 3 \times 2^{4} - 6 \times 2^{2} - 8 \times 2 - 2$
$\Rightarrow p (2) = 48 - 24 - 16 - 2$
$\Rightarrow p (2) = 6$ .

Hence, the remainder when $p (x)$ is divided by $g (x)$ is 6.

6. $p (x) = 2 x^{3} - 9 x^{2} + x + 15, g (x) = 2 x - 3$

Answer

We have
$p (x) = 2 x^{3} - 9 x^{2} + x + 15, g (x) = 2 x - 3$
Here, $g (x) = 0$
$\Rightarrow 2 x - 3 = 0$
$\Rightarrow x = \frac{3}{2}$

According to remainder theorem, when $p (x)$ is divided by $g (x)$ , the remainder is $p (\frac{3}{2})$ .

$∵ p (x) = 2 x^{3} - 9 x^{2} + x + 15$
$\Rightarrow p (\frac{3}{2}) = 2 {(\frac{3}{2})}^{3} - 9 {(\frac{3}{2})}^{2} + (\frac{3}{2}) + 15$
$\Rightarrow p (\frac{3}{2}) = 2 \times \frac{27}{8} - 9 \times \frac{9}{4} + \frac{3}{2} + 15$
$\Rightarrow p (\frac{3}{2}) = \frac{27}{4} - \frac{81}{4} + \frac{3}{2} + 15$
$\Rightarrow p (\frac{3}{2}) = \frac{27 - 81 + 6 + 60}{4} = \frac{12}{4} = 3$ .

Hence, the remainder when $p (x)$ is divided by $g (x)$ is 3.

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

Answer

We have
$p (x) = x^{3} - 2 x^{2} - 8 x - 1, g (x) = x + 1.$
Here, $g (x) = 0$
$\Rightarrow x + 1 = 0$
$\Rightarrow x = - 1$ .

According to remainder theorem, when $p (x)$ is divided by $g (x)$ , the remainder is $p (- 1)$ .

$∵ p (x) = x^{3} - 2 x^{2} - 8 x - 1$
$\Rightarrow p (- 1) = {(- 1)}^{3} - 2 \times {(- 1)}^{2} - 8 \times (- 1) - 1$
$\Rightarrow p (- 1) = - 1 - 2 + 8 - 1$
$\Rightarrow p (- 1) = 4$ .

Hence, the remainder when $p (x)$ is divided by $g (x)$ is 4.

8. $p (x) = 2 x^{3} + x^{2} - 15 x - 12, g (x) = x + 2.$

Answer

We have
$p (x) = 2 x^{3} + x^{2} - 15 x - 12, g (x) = x + 2.$
Here, $g (x) = 0$
$\Rightarrow x + 2 = 0$
$\Rightarrow x = - 2$ .

According to remainder theorem, when $p (x)$ is divided by $g (x)$ , the remainder is $p (- 2)$ .

$∵ p (x) = 2 x^{3} + x^{2} - 15 x - 12$
$\Rightarrow p (- 2) = 2 \times {(- 2)}^{3} + {(- 2)}^{2} - 15 \times (- 2) - 12$
$\Rightarrow p (- 2) = - 16 + 4 + 30 - 12$
$\Rightarrow p (- 2) = 6$ .

Hence, the remainder when $p (x)$ is divided by $g (x)$ is 6.

9. $p (x) = 6 x^{3} + 13 x^{2} + 3, g (x) = 3 x + 2.$

Answer

We have
$p (x) = 6 x^{3} + 13 x^{2} + 3, g (x) = 3 x + 2.$
Here, $g (x) = 0$
$\Rightarrow 3 x + 2 = 0$
$\Rightarrow x = \frac{- 2}{3}$ .

According to remainder theorem, when $p (x)$ is divided by $g (x)$ , the remainder is $p (\frac{- 2}{3})$ .

$∵ p (x) = 6 x^{3} + 13 x^{2} + 3$
$\Rightarrow p (\frac{- 2}{3}) = 6 \times {(\frac{- 2}{3})}^{3} + 13 \times {(\frac{- 2}{3})}^{2} + 3$
$\Rightarrow p (\frac{- 2}{3}) = 6 \times \frac{- 8}{27} + 13 \times \frac{4}{9} + 3$
$\Rightarrow p (\frac{- 2}{3}) = \frac{- 16}{9} + \frac{52}{9} + 3$
$\Rightarrow p (\frac{- 2}{3}) = \frac{- 16 + 52 + 27}{9}$
$\Rightarrow p (\frac{- 2}{3}) = \frac{63}{9}$
$\Rightarrow p (\frac{- 2}{3}) = 7$ .

Hence, the remainder when $p (x)$ is divided by $g (x)$ is 7.

10. $p (x) = x^{3} - 6 x^{2} + 2 x - 4, g (x) = 1 - \frac{3}{2} x .$

Answer

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

According to remainder theorem, when $p (x)$ is divided by $g (x)$ , the remainder is $p (\frac{2}{3})$ .

$∵ p (x) = x^{3} - 6 x^{2} + 2 x - 4$
$\Rightarrow p (\frac{2}{3}) = {(\frac{2}{3})}^{3} - 6 \times {(\frac{2}{3})}^{2} + 2 \times (\frac{2}{3}) - 4$
$\Rightarrow p (\frac{2}{3}) = \frac{8}{27} - 6 \times \frac{4}{9} + 2 \times \frac{2}{3} - 4$
$\Rightarrow p (\frac{2}{3}) = \frac{8}{27} - \frac{8}{3} + \frac{4}{3} - 4$
$\Rightarrow p (\frac{2}{3}) = \frac{8 - 72 + 36 - 108}{27}$
$\Rightarrow p (\frac{2}{3}) = \frac{- 136}{27}$ .

Hence, the remainder when $p (x)$ is divided by $g (x)$ is $\frac{- 136}{27}$ .

11. $p (x) = 2 x^{3} + 3 x^{2} - 11 x - 3, g (x) = (x + \frac{1}{2}) .$

Answer

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

According to remainder theorem, when $p (x)$ is divided by $g (x)$ , the remainder is $p (\frac{- 1}{2})$ .

$∵ p (x) = 2 x^{3} + 3 x^{2} - 11 x - 3$
$\Rightarrow p (- \frac{1}{2}) = 2 \times {(- \frac{1}{2})}^{3} + 3 \times {(- \frac{1}{2})}^{2} - 11 \times (- \frac{1}{2}) - 3$
$\Rightarrow p (- \frac{1}{2}) = 2 \times \frac{- 1}{8} + 3 \times \frac{1}{4} + 11 \times \frac{1}{2} - 3$
$\Rightarrow p (- \frac{1}{2}) = \frac{- 1}{4} + \frac{3}{4} + \frac{11}{2} - 3$
$\Rightarrow p (- \frac{1}{2}) = \frac{- 1 + 3 + 22 - 12}{4}$
$\Rightarrow p (- \frac{1}{2}) = \frac{12}{4}$
$\Rightarrow p (- \frac{1}{2}) = 3$

Hence, the remainder when $p (x)$ is divided by $g (x)$ is $3$ .

12. $p (x) = x^{3} - a x^{2} + 6 x - a, g (x) = x - a .$

Answer

We have
$p (x) = x^{3} - a x^{2} + 6 x - a, g (x) = x - a .$
Here, $g (x) = 0$
$\Rightarrow x - a = 0$
$\Rightarrow x = a$ .

According to remainder theorem, when $p (x)$ is divided by $g (x)$ , the remainder is $p (a)$ .

$∵ p (x) = x^{3} - a x^{2} + 6 x - a$
$\Rightarrow p (a) = a^{3} - a \times a^{2} + 6 a - a$
$\Rightarrow p (a) = a^{3} - a^{3} + 6 a - a$
$\Rightarrow x = 5 a$ .

Hence, the remainder when $p (x)$ is divided by $g (x)$ is $5 a$ .

In Exercise 2C RS Aggarwal Class 9, the thirteenth question is a word problem on remainder theorem.

13. The polynomials $(2 x^{3} + x^{2} - a x + 2)$ and $(2 x^{3} - 3 x^{2} - 3 x + a)$ when divided by $(x - 2)$ leave the same remainder. Find the value of a.

Answer

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

According to remainder theorem, when $p_{1} (x)$ and $p_{2} (x)$ are divided by $x - 2$ , the obtained remainders are $p_{1} (2)$ and $p_{2} (2)$ respectively.

Since the remainder obtained when $p_{1} (x)$ is divided by $g (x)$ is equal to the remainder obtained when $p_{2} (x)$ is divided by $g (x)$ .
$∴ p_{1} (2)$ $= p_{2} (2)$
$\Rightarrow 2 \times 2^{3} + 2^{2} - 2 a + 2 = 2 \times 2^{3} - 3 \times 2^{2} - 3 \times 2 + a$
$\Rightarrow 16 + 4 - 2 a + 2 = 16 - 12 - 6 + a$
$\Rightarrow - 2 a - a = - 12 - 6 - 2 - 4$
$\Rightarrow - 3 a = - 24$
$\Rightarrow a = 8$ .

Hence, the value of $a$ is 8.

In Exercise 2C RS Aggarwal Class 9, the fourteenth question is a word problem that has been comprehensively solved below.

14. The polynomial $p (x) = x^{4} - 2 x^{3} + 3 x^{2} - a x + b$ when divided by $(x - 1)$ and $(x + 1)$ leaves the remainders 5 and 19 respectively. Find the values of $a$ and $b$ . Hence, find the remainder when $p (x)$ is divided by $(x - 2)$ .

Answer

We have
$p (x) = x^{4} - 2 x^{3} + 3 x^{2} - a x + b$

$∵ x - 1 = 0$
$\Rightarrow x = 1$

According to remainder theorem, when $p (x)$ is divided by $(x - 1)$ , the remainder is $p (1)$ .
$∴ p (1) = 5$
$\Rightarrow 1^{4} - 2 \times 1^{3} + 3 \times 1^{2} - a \times 1 + b = 5$
$\Rightarrow 1 - 2 + 3 - a + b = 5$
$\Rightarrow 2 - a + b = 5$
$\Rightarrow - a + b = 5 - 2$
$\Rightarrow - a + b = 3$ . . . (1)

Also, when $p (x)$ is divided by $(x + 1)$ , the remainder is $p (- 1)$ .
$∴ p (- 1) = 19$
$\Rightarrow {(- 1)}^{4} - 2 \times {(- 1)}^{3} + 3 \times {(- 1)}^{2} - a \times (- 1) + b = 19$
$\Rightarrow 1 + 2 + 3 + a + b = 19$
$\Rightarrow a + b = 19 - 6$
$\Rightarrow a + b = 13$ . . . (2)

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

- a + b = 3

a + b = 13

2 b = 16

\Rightarrow b = 8

putting $b = 8$ in equation (2), we get.
$a + 8 = 13$
$\Rightarrow a = 13 - 8$
$\Rightarrow a = 5$

Putting the values of $a$ and $b$ in $p (x)$ , we get.
$p (x) = x^{4} - 2 x^{3} + 3 x^{2} - 5 x + 8$ .

Now, the remainder when $p (x)$ is divided by $(x - 2)$ is $p (2)$ .
$∴ p (2) = 2^{4} - 2 \times 2^{3} + 3 \times 2^{2} - 5 \times 2 + 8$
$\Rightarrow p (2) = 16 - 16 + 12 - 10 + 8$
$\Rightarrow p (2) = 10$ .

Hence, the remainder when $p (x)$ is divided by $(x - 2)$ is 10.

In Exercise 2C RS Aggarwal Class 9, the fifteenth question is a word problem and its comprehensive solution is given below.

15. If $p (x) = x^{3} - 5 x^{2} + 4 x - 3$ and $g (x) = (x - 2)$ show that $p (x)$ is not a multiple of $g (x)$ .

Answer

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

We know that $p (x)$ is a multiple of $g (x)$ if the remainder obtained is zero when $p (x)$ is divided by $g (x)$ .

$∵ g (x) = 0$
$\Rightarrow (x - 2) = 0$
$\Rightarrow x = 2$

According to remainder theorem, when $p (x)$ is divided by $g (x)$ , the remainder is $p (2)$ .

$∴ p (2) = 2^{3} - 5 \times 2^{2} + 4 \times 2 - 3$
$\Rightarrow p (2) = 8 - 20 + 8 - 3$
$\Rightarrow p (2) = 16 - 23$
$\Rightarrow p (2) = - 7$

∵ The remainder is not equal to zero, the given polynomial $p (x)$ is not a multiple of $g (x)$ .

In Exercise 2C RS Aggarwal Class 9, the sixteenth question is about the application of factor theorem.

16. If $p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5$ and $g (x) =$ $(2 x + 1)$ , show that $g (x)$ is not a factor of $p (x)$ .

Answer

We have
$p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5$
$g (x) =$ $(2 x + 1)$
Here, $g (x) = 0$
$\Rightarrow 2 x + 1 = 0$
$\Rightarrow x = \frac{- 1}{2}$ .

For $g (x)$ to be a factor of $p (x)$ , the remainder obtained when $p (x)$ is divided by $g (x)$ must be zero.

Putting $x = \frac{- 1}{2}$ in $p (x)$ , we get.

$p (\frac{- 1}{2}) = 2 {(\frac{- 1}{2})}^{3} - 11 {(\frac{- 1}{2})}^{2} - 4 \times (\frac{- 1}{2}) + 5$
$\Rightarrow p (\frac{- 1}{2}) = 2 \times \frac{- 1}{8} - 11 \times \frac{1}{4} - 4 \times \frac{- 1}{2} + 5$
$\Rightarrow p (\frac{- 1}{2}) = \frac{- 1}{4} - \frac{11}{4} + 2 + 5$
$\Rightarrow p (\frac{- 1}{2}) = \frac{- 1}{4} - \frac{11}{4} + 7$
$\Rightarrow p (\frac{- 1}{2}) = \frac{- 1 - 11 + 28}{4}$
$\Rightarrow p (\frac{- 1}{2}) = \frac{16}{4}$
$\Rightarrow p (\frac{- 1}{2}) = 4$

$∵ p (\frac{- 1}{2}) \neq 0$ , $g (x) = 2 x + 1$ is not a factor of $p (x)$ .

Hence, proved!

Exercise 2C RS Aggarwal Class 9 Mathematics Solutions

1. By actual division, find the quotient and the remainder when $(x^{4} + 1)$ is divided by $(x - 1)$ .
Verify that remainder = $f (1)$ .

Answer

2. Verify the division algorithm for the polynomials
$p (x) = 2 x^{4} - 6 x^{3} + 2 x^{2} - x + 2$ and $g (x) = x + 2$ .

Answer

Verification of Division Algorithm:

Using the remainder theorem, find the remainder, when $p (x)$ is divided by $g (x)$ , where

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

Answer

4. $p (x) = 2 x^{3} - 7 x^{2} + 9 x - 13, g (x) = x - 3$

Answer

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

Answer

6. $p (x) = 2 x^{3} - 9 x^{2} + x + 15, g (x) = 2 x - 3$

Answer

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

Answer

8. $p (x) = 2 x^{3} + x^{2} - 15 x - 12, g (x) = x + 2.$

Answer

9. $p (x) = 6 x^{3} + 13 x^{2} + 3, g (x) = 3 x + 2.$

Answer

10. $p (x) = x^{3} - 6 x^{2} + 2 x - 4, g (x) = 1 - \frac{3}{2} x .$

Answer

11. $p (x) = 2 x^{3} + 3 x^{2} - 11 x - 3, g (x) = (x + \frac{1}{2}) .$

Answer

12. $p (x) = x^{3} - a x^{2} + 6 x - a, g (x) = x - a .$

Answer

13. The polynomials $(2 x^{3} + x^{2} - a x + 2)$ and $(2 x^{3} - 3 x^{2} - 3 x + a)$ when divided by $(x - 2)$ leave the same remainder. Find the value of a.

Answer

14. The polynomial $p (x) = x^{4} - 2 x^{3} + 3 x^{2} - a x + b$ when divided by $(x - 1)$ and $(x + 1)$ leaves the remainders 5 and 19 respectively. Find the values of $a$ and $b$ . Hence, find the remainder when $p (x)$ is divided by $(x - 2)$ .

Answer

15. If $p (x) = x^{3} - 5 x^{2} + 4 x - 3$ and $g (x) = (x - 2)$ show that $p (x)$ is not a multiple of $g (x)$ .

Answer

16. If $p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5$ and $g (x) =$ $(2 x + 1)$ , show that $g (x)$ is not a factor of $p (x)$ .

Answer

Important Links

CBSE Class 9 Sample Papers 2025-26

Download NCERT Book Class 9th

Buy RS Aggarwal Class 9 Book

Related Links 🔗

Exercise 2C RS Aggarwal Class 9 Mathematics Solutions

1. By actual division, find the quotient and the remainder when (x4+1) is divided by (x−1). Verify that remainder = f(1).

Answer

2. Verify the division algorithm for the polynomialsp(x)=2x4−6x3+2x2−x+2 and g(x)=x+2.

Answer

Verification of Division Algorithm:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where

3. p(x)=x3−6x2+9x+3, g(x)=x−1

Answer

4. p(x)=2x3−7x2+9x−13, g(x)=x−3

Answer

5. p(x)=3x4−6x2−8x−2, g(x)=x−2

Answer

6. p(x)=2x3−9x2+x+15, g(x)=2x−3

Answer

7. p(x)=x3−2x2−8x−1, g(x)=x+1.

Answer

8. p(x)=2x3+x2−15x−12, g(x)=x+2.

Answer

9. p(x)=6x3+13x2+3, g(x)=3x+2.

Answer

10. p(x)=x3−6x2+2x−4, g(x)=1−32x.

Answer

11. p(x)=2x3+3x2−11x−3, g(x)=(x+12).

Answer

12. p(x)=x3−ax2+6x−a, g(x)=x−a.

Answer

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

Answer

14. The polynomial p(x)=x4−2x3+3x2−ax+b when divided by (x−1) and (x+1) leaves the remainders 5 and 19 respectively. Find the values of a and b. Hence, find the remainder when p(x) is divided by (x−2).

Answer

15. If p(x)=x3−5x2+4x−3 and g(x)=(x−2) show that p(x) is not a multiple of g(x).

Answer

16. If p(x)=2x3−11x2−4x+5 and g(x)= (2x+1), show that g(x) is not a factor of p(x).

Answer

Important Links

CBSE Class 9 Sample Papers 2025-26

Download NCERT Book Class 9th

Buy RS Aggarwal Class 9 Book

Related Links 🔗

1. By actual division, find the quotient and the remainder when $(x^{4} + 1)$ is divided by $(x - 1)$ .
Verify that remainder = $f (1)$ .

2. Verify the division algorithm for the polynomials
$p (x) = 2 x^{4} - 6 x^{3} + 2 x^{2} - x + 2$ and $g (x) = x + 2$ .

Using the remainder theorem, find the remainder, when $p (x)$ is divided by $g (x)$ , where

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

4. $p (x) = 2 x^{3} - 7 x^{2} + 9 x - 13, g (x) = x - 3$

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

6. $p (x) = 2 x^{3} - 9 x^{2} + x + 15, g (x) = 2 x - 3$

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

8. $p (x) = 2 x^{3} + x^{2} - 15 x - 12, g (x) = x + 2.$

9. $p (x) = 6 x^{3} + 13 x^{2} + 3, g (x) = 3 x + 2.$

10. $p (x) = x^{3} - 6 x^{2} + 2 x - 4, g (x) = 1 - \frac{3}{2} x .$

11. $p (x) = 2 x^{3} + 3 x^{2} - 11 x - 3, g (x) = (x + \frac{1}{2}) .$

12. $p (x) = x^{3} - a x^{2} + 6 x - a, g (x) = x - a .$

13. The polynomials $(2 x^{3} + x^{2} - a x + 2)$ and $(2 x^{3} - 3 x^{2} - 3 x + a)$ when divided by $(x - 2)$ leave the same remainder. Find the value of a.

14. The polynomial $p (x) = x^{4} - 2 x^{3} + 3 x^{2} - a x + b$ when divided by $(x - 1)$ and $(x + 1)$ leaves the remainders 5 and 19 respectively. Find the values of $a$ and $b$ . Hence, find the remainder when $p (x)$ is divided by $(x - 2)$ .

15. If $p (x) = x^{3} - 5 x^{2} + 4 x - 3$ and $g (x) = (x - 2)$ show that $p (x)$ is not a multiple of $g (x)$ .

16. If $p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5$ and $g (x) =$ $(2 x + 1)$ , show that $g (x)$ is not a factor of $p (x)$ .