RS AGGARWAL CLASS 9 Chapter 2 POLYNOMIALS EXERCISE 2C

 EXERCISE 2C

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Question 1:

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

Answer 1:

Let f(x) = x⁴ + 1 and g(x) = x – 1.



Quotient = x³ + x² + x + 1

Remainder = 2

Verification:

Putting x = 1 in f(x), we get

f(1) = 1⁴ + 1 = 1 + 1 = 2 = Remainder, when f(x) = x⁴ + 1 is divided by g(x) = x – 1

Question 2:

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

Answer 2:

$p\left(x\right)=2x^4-6x^3+2x^2-x+2$$p\left(x\right)=2x^4-6x^3+2x^2-x+2$ and $g\left(x\right)=x+2$



Quotient = $2x^3-10x^2+22x-45$$2x^3-10x^2+22x-45$

Remainder = 92

Verification:

Divisor × Quotient + Remainder
$=\left(x+2\right)\times \left(2x^3-10x^2+22x-45\right)+92=x\left(2x^3-10x^2+22x-45\right)+2\left(2x^3-10x^2+22x-45\right)+92=2x^4-10x^3+22x^2-45x+4x^3-20x^2+44x-90+92=2x^4-6x^3+2x^2-x+2$$$\begin{gathered} =\left(x+2\right)\times \left(2x^3-10x^2+22x-45\right)+92 \\ =x\left(2x^3-10x^2+22x-45\right)+2\left(2x^3-10x^2+22x-45\right)+92 \\ =2x^4-10x^3+22x^2-45x+4x^3-20x^2+44x-90+92 \\ =2x^4-6x^3+2x^2-x+2 \end{gathered}$$
= Dividend

Hence verified.

Question 3:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p\left(x\right)=x^3-6x^2+9x+3, g\left(x\right)=x-1$.

Answer 3:

$p\left(x\right)=x^3-6x^2+9x+3$

​$g\left(x\right)=x-1$

By remainder theorem, when p(x) is divided by (x − 1), then the remainder = p(1).

Putting x = 1 in p(x), we get

$p\left(1\right)=1^3-6\times 1^2+9\times 1+3=1-6+9+3=7$ 

∴ Remainder = 7

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

Question 4:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p\left(x\right)=2x^3-7x^2+9x-13, g\left(x\right)=x-3$.

Answer 4:

$p\left(x\right)=2x^3-7x^2+9x-13$

​$g\left(x\right)=x-3$

By remainder theorem, when p(x) is divided by (x − 3), then the remainder = p(3).

Putting x = 3 in p(x), we get

$p\left(3\right)=2\times 3^3-7\times 3^2+9\times 3-13=54-63+27-13=5$ 

∴ Remainder = 5

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

Question 5:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p\left(x\right)=3x^4-6x^2-8x-2, g\left(x\right)=x-2$.

Answer 5:

$p\left(x\right)=3x^4-6x^2-8x-2$

​$g\left(x\right)=x-2$

By remainder theorem, when p(x) is divided by (x − 2), then the remainder = p(2).

Putting x = 2 in p(x), we get

$p\left(2\right)=3\times 2^4-6\times 2^2-8\times 2-2=48-24-16-2=6$ 

∴ Remainder = 6

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

Question 6:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p\left(x\right)=2x^3-9x^2+x+15, g\left(x\right)=2x-3$.

Answer 6:

$p\left(x\right)=2x^3-9x^2+x+15$

​$g\left(x\right)=2x-3=2\left(x-\frac{3}{2}\right)$

By remainder theorem, when p(x) is divided by (2x − 3), then the remainder = $p\left(\frac{3}{2}\right)$.

Putting $x=\frac{3}{2}$ in p(x), we get

$p\left(\frac{3}{2}\right)=2\times {\left(\frac{3}{2}\right)}^3-9\times {\left(\frac{3}{2}\right)}^2+\frac{3}{2}+15=\frac{27}{4}-\frac{81}{4}+\frac{3}{2}+15$ $=\frac{27-81+6+60}{4}=\frac{12}{4}=3$

∴ Remainder = 3

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

Question 7:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p\left(x\right)=x^3-2x^2-8x-1, g\left(x\right)=x+1$.

Answer 7:

$p\left(x\right)=x^3-2x^2-8x-1$

​$g\left(x\right)=x+1$

By remainder theorem, when p(x) is divided by (x + 1), then the remainder = p(−1).

Putting x = −1 in p(x), we get

$p\left(-1\right)={\left(-1\right)}^3-2\times {\left(-1\right)}^2-8\times \left(-1\right)-1=-1-2+8-1=4$ 

∴ Remainder = 4

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

Question 8:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p\left(x\right)=2x^3+x^2-15x-12, g\left(x\right)=x+2$.

Answer 8:

$p\left(x\right)=2x^3+x^2-15x-12$

​$g\left(x\right)=x+2$

By remainder theorem, when p(x) is divided by (x + 2), then the remainder = p(−2).

Putting x = −2 in p(x), we get

$p\left(-2\right)=2\times {\left(-2\right)}^3+{\left(-2\right)}^2-15\times \left(-2\right)-12=-16+4+30-12=6$ 

∴ Remainder = 6

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

Question 9:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p\left(x\right)=6x^3+13x^2+3, g\left(x\right)=3x+2$.

Answer 9:

$p\left(x\right)=6x^3+13x^2+3$

​$g\left(x\right)=3x+2=3\left(x+\frac{2}{3}\right)=3\left[x-\left(-\frac{2}{3}\right)\right]$

By remainder theorem, when p(x) is divided by (3x + 2), then the remainder = $p\left(-\frac{2}{3}\right)$.

Putting $x=-\frac{2}{3}$ in p(x), we get

$p\left(-\frac{2}{3}\right)=6\times {\left(-\frac{2}{3}\right)}^3+13\times {\left(-\frac{2}{3}\right)}^2+3=-\frac{16}{9}+\frac{52}{9}+3$ $=\frac{-16+52+27}{9}=\frac{63}{9}=7$

∴ Remainder = 7

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

Question 10:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p\left(x\right)=x^3-6x^2+2x-4, g\left(x\right)=1-\frac{3}{2}x$.

Answer 10:

$p\left(x\right)=x^3-6x^2+2x-4$

​$g\left(x\right)=1-\frac{3}{2}x=-\frac{3}{2}\left(x-\frac{2}{3}\right)$

By remainder theorem, when p(x) is divided by $\left(1-\frac{3}{2}x\right)$, then the remainder = $p\left(\frac{2}{3}\right)$.

Putting $x=\frac{2}{3}$ in p(x), we get

$p\left(\frac{2}{3}\right)={\left(\frac{2}{3}\right)}^3-6\times {\left(\frac{2}{3}\right)}^2+2\times \left(\frac{2}{3}\right)-4=\frac{8}{27}-\frac{8}{3}+\frac{4}{3}-4$ $=\frac{8-72+36-108}{27}=-\frac{136}{27}$

∴ Remainder = $-\frac{136}{27}$

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

Question 11:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p\left(x\right)=2x^3+3x^2-11x-3, g\left(x\right)=\left(x+\frac{1}{2}\right)$.

Answer 11:

$p\left(x\right)=2x^3+3x^2-11x-3$

​$g\left(x\right)=\left(x+\frac{1}{2}\right)=\left[x-\left(-\frac{1}{2}\right)\right]$

By remainder theorem, when p(x) is divided by $\left(x+\frac{1}{2}\right)$, then the remainder = $p\left(-\frac{1}{2}\right)$.

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

$p\left(-\frac{1}{2}\right)=2\times {\left(-\frac{1}{2}\right)}^3+3\times {\left(-\frac{1}{2}\right)}^2-11\times \left(-\frac{1}{2}\right)-3=-\frac{1}{4}+\frac{3}{4}+\frac{11}{2}-3$ $=\frac{-1+3+22-12}{4}=\frac{12}{4}=3$

∴ Remainder = 3

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

Question 12:

Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$p\left(x\right)=x^3-a{x}^2+6x-a, g\left(x\right)=x-a$.

Answer 12:

$p\left(x\right)=x^3-a{x}^2+6x-a$

​$g\left(x\right)=x-a$

By remainder theorem, when p(x) is divided by (x − a), then the remainder = p(a).

Putting x = a in p(x), we get

$p\left(a\right)=a^3-a\times a^2+6\times a-a=a^3-a^3+6a-a=5a$ 

∴ Remainder = 5a

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

Question 13:

The polynomials $\left(2x^3+x^2-ax+2\right) \mathrm{and} \left(2x^3-3x^2-3x+a\right)$ when divided by (x – 2) leave the same remainder. Find the value of a.

Answer 13:

Let $f\left(x\right)=2x^3+x^2-ax+2$ and $g\left(x\right)=2x^3-3x^2-3x+a$.

By remainder theorem, when f(x) is divided by (x – 2), then the remainder = f(2).

Putting x = 2 in f(x), we get

$f\left(2\right)=2\times 2^3+2^2-a\times 2+2=16+4-2a+2=-2a+22$ 

By remainder theorem, when g(x) is divided by (x – 2), then the remainder = g(2).

Putting x = 2 in g(x), we get

$g\left(2\right)=2\times 2^3-3\times 2^2-3\times 2+a=16-12-6+a=-2+a$

It is given that,

$$\begin{gathered} f\left(2\right)=g\left(2\right) \\ \Rightarrow -2a+22=-2+a \\ \Rightarrow -3a=-24 \\ \Rightarrow a=8 \end{gathered}$$
Thus, the value of a is 8.

Question 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 14:

 


Thus, we have:

And,


Now,





By putting the value of b, we get the value of a, i.e., 5.
∴ a = 5 and b = 8
Now,
$f\left(x\right) = x^4-2x^3 + 3x^2 - 5x + 8$
Also,

Thus, we have:

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Question 15:

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

Answer 15:

$p\left(x\right)=x^3-5x^2+4x-3$$p\left(x\right)=x^3-5x^2+4x-3$

$g\left(x\right)=x-2$$g\left(x\right)=x-2$

Putting x = 2 in p(x), we get

$p\left(2\right)=2^3-5\times 2^2+4\times 2-3=8-20+8-3=-7\ne 0$$p\left(2\right)=2^3-5\times 2^2+4\times 2-3=8-20+8-3=-7\ne 0$

Therefore, by factor theorem, (x − 2) is not a factor of p(x).

Hence, p(x) is not a multiple of g(x).

Question 16:

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

Answer 16:

$p\left(x\right)=2x^3-11x^2-4x+5$$p\left(x\right)=2x^3-11x^2-4x+5$

$g\left(x\right)=2x+1=2\left(x+\frac{1}{2}\right)=2\left[x-\left(-\frac{1}{2}\right)\right]$$g\left(x\right)=2x+1=2\left(x+\frac{1}{2}\right)=2\left[x-\left(-\frac{1}{2}\right)\right]$

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

$p\left(-\frac{1}{2}\right)=2\times {\left(-\frac{1}{2}\right)}^3-11\times {\left(-\frac{1}{2}\right)}^2-4\times \left(-\frac{1}{2}\right)+5=-\frac{1}{4}-\frac{11}{4}+2+5$$p\left(-\frac{1}{2}\right)=2\times {\left(-\frac{1}{2}\right)}^3-11\times {\left(-\frac{1}{2}\right)}^2-4\times \left(-\frac{1}{2}\right)+5=-\frac{1}{4}-\frac{11}{4}+2+5$ $=-\frac{12}{4}+7=-3+7=4\ne 0$$=-\frac{12}{4}+7=-3+7=4\ne 0$

Therefore, by factor theorem, (2x + 1) is not a factor of p(x).

Hence, g(x) is not a factor of p(x).