# RD Sharma Class 9 Factorisation of Polynomials Exercise 6.3 Solutions

On this page you will find Maths RD Sharma Class 9 Factorisation of Polynomials Exercise 6.3 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 6 for class 9 deals with the topic of triangles and it is one of the most important chapters.

## Download RD Sharma Class 9 Factorisation of Polynomials Exercise 6.3 Solutions in PDF

Students can use these solutions to overcome the fear of maths and the solutions have been designed in such a way that it enables them to discover easy ways to solve different problems. These solutions can help students in refining their maths fluency and problem-solving skills. Students can go through the RD Sharma solutions for class 9 chapter 6 Factorisation of Polynomials Exercise 6.3 below and it will be beneficial for them.

## RD Sharma Class 9 Factorisation of Polynomials Exercise 6.3 Solutions

In each of the following, using the remainder theorem, find the remainder when $f(x)$ is divided by $g(x)$ and verify the by actual division: $(1-8)$

01. $f(x)=x^{3}+4 x^{2}-3 x+10, g(x)=x+4$
Sol:
Here, $f(x)=x^{3}+4 x^{2}-3 x+10$
$g(x)=x+4$
from, the remainder theorem when $f(x)$ is divided by $g(x)=x-(-4)$ the remainder will be equal to $f(-4)$
Let, $g(x)=0$
$\Rightarrow x+4=0$
$\Rightarrow x=-4$
Substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f(-4)=(-4)^{3}+4(-4)^{2}-3(-4)+10$
$=-64+(4 * 16)+12+10$
$=-64+64+12+10$
$=12+10$
$=22$
Therefore, the remainder is 22

Q2. $f(x)=4 x^{4}-3 x^{3}-2 x^{2}+x-7, g(x)=x-1$
Sol:
Here, $f(x)=4 x^{4}-3 x^{3}-2 x^{2}+x-7$
$g(x)=x-1$
from, the remainder theorem when $f(x)$ is divided by $g(x)=x-(-1)$ the remainder will be equal to $f(1)$
Let,$g(x)=0$
$\Rightarrow x-1=0$
$\Rightarrow x=1$
Substitute the value of $x$ in $f(x)$
$f(1)=4(1)^{4}-3(1)^{3}-2(1)^{2}+1-7$
$=4-3-2+1-7$
$=5-12$
$=-7$
Therefore, the remainder is 7

Q3. $f(x)=2 x^{4}-6 x^{3}+2 x^{2}-x+2, g(x)=x+2$
Sol:
Here, $f(x)=2 x^{4}-6 x^{3}+2 x^{2}-x+2$
$g(x)=x+2$
from, the remainder theorem when $\mathrm{f}(\mathrm{x})$ is divided by $\mathrm{g}(\mathrm{x})=\mathrm{x}-(-2)$ the remainder will be equal to $\mathrm{f}(-2)$
Let,$g(x)=0$
$\Rightarrow x+2=0$
$\Rightarrow x=-2$
Substitute the value of $x$ in $f(x)$
$f(-2)=2(-2)^{4}-6(-2)^{3}+2(-2)^{2}-(-2)+2$
$=(2 * 16)-(6 *(-8))+(2 * 4)+2+2$
$=32+48+8+2+2$
$=92$
Therefore, the remainder is 92

Q4. $f(x)=4 x^{3}-12 x^{2}+14 x-3, g(x)=2 x-1$
Sol:
Here, $f(x)=4 x^{3}-12 x^{2}+14 x-3$
$g(x)=2 x-1$
from, the remainder theorem when $f(x)$ is divided by $g(x)=2\left(x-\frac{1}{2}\right)$, the remainder is equal to $f\left(\frac{1}{2}\right)$
Let,$g(x)=0$
$\Rightarrow 2 x-1=0$
$\Rightarrow 2 x=1$
$\Rightarrow x=\frac{1}{2}$
Substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f\left(\frac{1}{2}\right)=4\left(\frac{1}{2}\right)^{3}-12\left(\frac{1}{2}\right)^{2}+14\left(\frac{1}{2}-3\right.$
$=4\left(\frac{1}{8}\right)-12\left(\frac{1}{4}\right)+4\left(\frac{1}{2}\right)-3$
$=\left(\frac{1}{2}\right)-3+7-3$
$=\left(\frac{1}{2}\right)+1$
Taking L.C.M
$=\left(\frac{2+1}{2}\right)$
$=\left(\frac{3}{2}\right)$
Therefore, the remainder is $\left(\frac{3}{2}\right)$

Q5. $f(x)=x^{3}-6 x^{2}+2 x-4, g(x)=1-2 x$
Sol:
Here, $f(x)=x^{3}-6 x^{2}+2 x-4$
$g(x)=1-2 x$
from, the remainder theorem when $f(x)$ is divided by $g(x)=-2\left(x-\frac{1}{2}\right)$, the remainder is equal to $f\left(\frac{1}{2}\right)$
Let,$g(x)=0$
$\Rightarrow 1-2 x=0$
$\Rightarrow-2 x=-1$
$\Rightarrow 2 x=1$
$\Rightarrow x=\frac{1}{2}$
Substitute the value of $x$ in $f(x)$
$f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{3}-6\left(\frac{1}{2}\right)^{2}+2\left(\frac{1}{2}\right)-4$
$=\frac{1}{8}-8\left(\frac{1}{4}\right)+2\left(\frac{1}{2}\right)-4$
$=\frac{1}{8}-\left(\frac{1}{2}\right)+1-4$
$=\frac{1}{8}-\left(\frac{1}{2}\right)-3$
Taking L.C.M
$=\frac{1-4+8-32}{8}$
$=\frac{1-36}{8}$
$=\frac{1-36}{8}$
$=\frac{-35}{8}$
Therefore, the remainder is $\frac{-35}{8}$

Q6. $f(x)=x^{4}-3 x^{2}+4, g(x)=x-2$
Sol:
Here, $f(x)=x^{4}-3 x^{2}+4$
$g(x)=x-2$
from, the remainder theorem when $f(x)$ is divided by $g(x)=x-2$ the remainder will be equal to $f(2)$
let, $g(x)=0$
$\Rightarrow x-2=0$
$\Rightarrow x=2$
Substitute the value of $x$ in $f(x)$
$f(2)=2^{4}-3(2)^{2}+4$
$=16-\left(3^{*} 4\right)+4$
$=16-12+4$
$=20-12$
$=8$
Therefore, the remainder is 8

Q7. $f(x)=9 x^{3}-3 x^{2}+x-5, g(x)=x-\frac{2}{3}$
Sol:
Here, $f(x)=9 x^{3}-3 x^{2}+x-5$
$g(x)=x-\frac{2}{3}$
from, the remainder theorem when $f(x)$ is divided by $g(x)=x-\frac{2}{3}$ the remainder will be equal to $f\left(\frac{2}{3}\right)$
substitute the value of $x$ in $f(x)$
$f\left(\frac{2}{3}\right)=9\left(\frac{2}{3}\right)-3\left(\frac{2}{3}\right)^{2}+\left(\frac{2}{3}\right)-5$
$=9\left(\frac{8}{27}\right)-3\left(\frac{4}{9}\right)+\frac{2}{3}-5$
$=\left(\frac{8}{3}\right)-\left(\frac{4}{3}\right)+\frac{2}{3}-5$
$=\frac{8-4+2-15}{3}$
$=\frac{10-19}{3}$
$=\frac{-9}{3}$
$=-3$
Therefore, the remainder is $-3$

Q8. $f(x)=3 x^{4}+2 x^{3}-\frac{x^{3}}{3}-\frac{x}{9}+\frac{2}{27}, g(x)=x+\frac{2}{3}$
Sol:
Here, $f(x)=3 x^{4}+2 x^{3}-\frac{x^{3}}{3}-\frac{x}{9}+\frac{2}{27}$
$g(x)=x+\frac{2}{3}$
from remainder theorem when $f(x)$ is divided by $g(x)=x-\left(-\frac{2}{3}\right)$, the remainder is equal to $f\left(-\frac{2}{3}\right)$
substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f\left(-\frac{2}{3}\right)=3\left(-\frac{2}{3}\right)^{4}+2\left(-\frac{2}{3}\right)^{3}-\frac{\left(-\frac{2}{3}\right)^{3}}{3}$ $-(\operatorname{lfrac}\{(-\operatorname{lfrac}\{2\}\{3\})\}\{9\}+[$ latex $] \operatorname{frac}\{2\}\{27\}(\operatorname{lfrac}\{2\}\{27\})]^{\prime \prime}>$
$=[$ latex $] 3( \operatorname\{16\}\{81\})+2(\operatorname\{-8\}\{27\})- \operatorname\{4\}\{(9 \star 3)\}-(\operatorname\{-2\}\{(9 * 3)\})+\operatorname\{2\}\{27\})$
$=\left(\frac{16}{27}\right)-\left(\frac{16}{27}\right)-\frac{4}{27}+\left(\frac{2}{27}\right)+\frac{2}{27}$
$=\left(\frac{4}{27}\right)-\left(\frac{4}{27}\right)$
$=0$
Therefore, the remainder is 0

Q9. If the polynomial $2 \mathrm{x}^{3}+\mathrm{ax}^{2}+3 \mathrm{x}-5$ and $\mathrm{x}^{3}+\mathrm{x}^{2}-4 \mathrm{x}+\mathrm{a}$ leave the same remainder when divided by $\mathrm{x}-2$, Find the value of a
Sol:
Given , the polymials are
$f(x)=2 x^{3}+a x^{2}+3 x-5$
$p(x)=x^{3}+x^{2}-4 x+a$
The remainders are $f(2)$ and $p(2)$ when $f(x)$ and $p(x)$ are divided by $x-2$
We know that,
$f(2)=p(2) \quad$ (given in problem)
we need to calculate $f(2)$ and $p(2)$
for, $f(2)$
substitute $(x=2)$ in $f(x)$
$f(2)=2(2)^{3}+a(2)^{2}+3(2)-5$
$=(2 * 8)+a 4+6-5$
$=16+4 a+1$
$=4 a+17—1$
for, $p(2)$
substitute $(x=2)$ in $p(x)$
$p(2)=2^{3}+2^{2}-4(2)+a$
$=8+4-8+a$
$=4+a—2$
Since, $f(2)=p(2)$
Equate eqn 1 and 2
$\Rightarrow 4 a+17=4+a$
$\Rightarrow 4 a-a=4-17$
$\Rightarrow 3 a=-13$
$\Rightarrow a=\frac{-13}{3}$
The value of $\mathrm{a}=\frac{-13}{3}$

Q10. If polynomials $a x^{3}+3 x^{2}-3$ and $2 x^{3}-5 x+a$ when divided by $(x-4)$ leave the remainders as $R_{1}$ and $R_{2}$ respectively. Find the values of a in each of the following cases, if
1. $\mathrm{R}_{1}=\mathrm{R}_{2}$
2. $\mathrm{R}_{1}+\mathrm{R}_{2}=0$
3. $2 \mathrm{R}_{1}-\mathrm{R}_{2}=0$
Sol:
Here, the polynomials are
$f(x)=a x^{3}+3 x^{2}-3$
$p(x)=2 x^{3}-5 x+a$
let,
$\mathrm{R}_{1}$ is the remainder when $\mathrm{f}(\mathrm{x})$ is divided by $\mathrm{x}-4$
$\Rightarrow \mathrm{R}_{1}=\mathrm{f}(4)$
$\Rightarrow \mathrm{R}_{1}=\mathrm{a}(4)^{3}+3(4)^{2}-3$
$=64 a+48-3$
$=64 a+45 \quad—1$
Now, let
$\mathrm{R}_{2}$ is the remainder when $\mathrm{p}(\mathrm{x})$ is divided by $\mathrm{x}-4$
$\Rightarrow \mathrm{R}_{2}=\mathrm{p}(4)$
$\Rightarrow \mathrm{R}_{2}=2(4)^{3}-5(4)+\mathrm{a}$
$=128-20+a$
$=108+a—2$
1. Given, $\mathrm{R}_{1}=\mathrm{R}_{2}$
$\Rightarrow 64 a+45=108+a$
$\Rightarrow 63 a=63$
$\Rightarrow a=1$
2. Given, $\mathrm{R}_{1}+\mathrm{R}_{2}=0$
$\Rightarrow 64 a+45+108+a=0$
$\Rightarrow 65 a+153=0$
$\Rightarrow a=\frac{-153}{65}$
3.Given, $2 \mathrm{R}_{1}-\mathrm{R}_{2}=0$
$\Rightarrow 2(64 a+45)-108-a=0$
$\Rightarrow>128 a+90-108-a=0$
$\Rightarrow 127 a-18=0$
$\Rightarrow a=\frac{18}{127}$

Q11. If the polynomials $a x^{3}+3 x^{2}-13$ and $2 x^{3}-5 x+$ a when divided by $(x-2)$ leave the same remainder, Find the value of a
Sol:
Here, the polynomials are
$f(x)=a x^{3}+3 x^{2}-13$
$p(x)=2 x^{3}-5 x+a$
equate, $x-2=0$
$x=2$
substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$ and $\mathrm{p}(\mathrm{x})$
$f(2)=(2)^{3}+3(2)^{2}-13$
$=8 a+12-13$
$=8 a-1$ ———1
$p(2)=2(2)^{3}-5(2)+a$
$=16-10+a$
$=6+a \quad—-2$
$f(2)=p(2)$
$\Rightarrow 8 a-1=6+a$
$\Rightarrow 8 a-a=6+1$
$\Rightarrow \quad 7 a=7$
$\Rightarrow \quad a=1$
The value of $a=1$

Q12. Find the remainder when $\mathrm{x}^{3}+3 \mathrm{x}^{3}+3 \mathrm{x}+1$ is divided by,
1. $x+1$
2. $x-\frac{1}{2}$
3. $x$
4. $x+\pi$
5. $5+2 \mathrm{x}$
Sol:
Here, $f(x)=x^{3}+3 x^{2}+3 x+1$
by remainder theorem
1. $\Rightarrow x+1=0$
$\Rightarrow x=-1$
substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f(-1)=(-1)^{3}+3(-1)^{2}+3(-1)+1$
$=-1+3-3+1$
$=0$
2. $x-\frac{1}{2}$
Sol:
Here, $f(x)=x^{3}+3 x^{2}+3 x+1$
By remainder theorem
$\Rightarrow x-\frac{1}{2}=0$
$\Rightarrow x=\frac{1}{2}$
substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{3}+3\left(\frac{1}{2}\right)^{2}+3\left(\frac{1}{2}\right)+1$
$=\left(\frac{1}{2}\right)^{3}+3\left(\frac{1}{2}\right)^{2}+3\left(\frac{1}{2}\right)+1$
$=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1$
$=\frac{1+6+12+8}{8}$
$=\frac{27}{8}$
3. $x$
Sol:
Here, $f(x)=x^{3}+3 x^{2}+3 x+1$
by remainder theorem
$\Rightarrow x=0$
substitute the value of $x$ in $f(x)$
$f(0)=0^{3}+3(0)^{2}+3(0)+1$
$=0+0+0+1$
$=1$
4. $x+\pi$
Sol:
Here, $f(x)=x^{3}+3 x^{2}+3 x+1$
by remainder theorem
$\Rightarrow x+\pi=0$
$\Rightarrow \quad x=-\pi$
Substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f(-\pi)=(-\pi)^{3}+3(-\pi)^{2}+3(-\pi)+1$
$=-(\pi)^{3}+3(\pi)^{2}-3(\pi)+1$
5. $5+2 x$
Sol:
Here, $f(x)=x^{3}+3 x^{2}+3 x+1$
by remainder theorem
$5+2 x=0$
$2 x=-5$
$x=\frac{-5}{2}$
substitute the value of $\mathrm{x}$ in $\mathrm{f}(\mathrm{x})$
$f\left(\frac{-5}{2}\right)=\left(\frac{-5}{2}\right)^{3}+3\left(\frac{-5}{2}\right)^{2}+3\left(\frac{-5}{2}\right)+1$
$=\frac{-125}{8}+3\left(\frac{25}{4}\right)+3\left(\frac{-5}{2}\right)+1$
$=\frac{-125}{8}+\frac{75}{4}-\frac{15}{2}+1$
Taking L.C.M
$=\frac{-125+150-50+8}{8}$
$=\frac{-27}{8}$

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