RD Sharma Class 9 Factorisation of Polynomials Exercise 6.2 Solutions

On this page you will find Maths RD Sharma Class 9 Factorisation of Polynomials Exercise 6.2 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.

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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 6.2 below and it will be beneficial for them.

RD Sharma Class 9 Factorisation of Polynomials Exercise 6.2 Solutions

Q1. If $\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-13 \mathrm{x}^{2}+17 \mathrm{x}+12$, Find
1. $f(2)$
2. $f(-3)$
3. $f(0)$
Sol:
The given polynomial is $\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-13 \mathrm{x}^{2}+17 \mathrm{x}+12$
1. $f(2)$
we need to substitute the ‘ 2 ‘ in $\mathrm{f}(\mathrm{x})$
$\mathrm{f}(2)=2(2)^{3}-13(2)^{2}+17(2)+12$
$=(2 * 8)-(13 * 4)+(17 * 2)+12$
$=16-52+34+12$
$=10$
therefore $f(2)=10$
2. $f(-3)$
we need to substitute the $^{\prime}(-3)$ ‘ in $f(x)$
$f(-3)=2(-3)^{3}-13(-3)^{2}+17(-3)+12$
$=(2 *-27)-(13 * 9)-(17 * 3)+12$
$=-54-117-51+12$
$=-210$
therefore $f(-3)=-210$
3. $f(0)$
we need to substitute the $^{\prime}(0)^{\prime}$ in $f(x)$
$\mathrm{f}(0)=2(0)^{3}-13(0)^{2}+17(0)+12$
$=(2 * 0)-(13 * 0)+(17 * 0)+12$
$=0-0+0+12$
$=12$
therefore $f(0)=12$

Q2. Verify whether the indicated numbers are zeros of the polynomial corresponding to them in the following cases:
1. $\mathrm{f}(\mathrm{x})=3 \mathrm{x}+1, \mathrm{x}=\frac{-1}{3}$
2. $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-1, \mathrm{x}=(1,-1)$
3. $\mathrm{g}(\mathrm{x})=3 \mathrm{x}^{2}-2, \mathrm{x}=\left(\frac{2}{\sqrt{3}}, \frac{-2}{\sqrt{3}}\right)$
4. $\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-6 \mathrm{x}^{2}+11 \mathrm{x}-6, \mathrm{x}=\mathbf{1}, \mathbf{2}, \mathbf{3}$
5. $\mathrm{f}(\mathrm{x})=5 \mathrm{x}-\pi, \mathrm{x}=\frac{4}{5}$
6. $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}, \mathbf{x}=\mathbf{0}$
7. $\mathrm{f}(\mathrm{x})=\mathrm{lx}+\mathrm{m}, \mathrm{x}=\frac{-\mathrm{m}}{\mathrm{l}}$
8. $\mathrm{f}(\mathrm{x})=2 \mathrm{x}+1, \mathrm{x}=\frac{1}{2}$
Sol:
$(1) \mathrm{f}(\mathrm{x})=3 \mathrm{x}+1, \mathrm{x}=\frac{-1}{3}$
we know that,
$f(x)=3 x+1$
substitute $x=\frac{-1}{3}$ in $f(x)$
$f\left(\frac{-1}{3}\right)=3\left(\frac{-1}{3}\right)+1$
$=-1+1$
$=0$
Since, the result is $0 x=\frac{-1}{3}$ is the root of $3 x+1$
(2) $f(x)=x^{2}-1, x=(1,-1)$
we know that,
$f(x)=x^{2}-1$
Given that $x=(1,-1)$
substitute $x=1$ in $f(x)$
$f(1)=1^{2}-1$
$=1-1$
$=0$
Now, substitute $x=(-1)$ in $f(x)$
$f(-1)=(-1)^{2}-1$
$=1-1$
$=0$
Since, the results when $x=(1,-1)$ are 0 they are the roots of the polynomial $f(x)=x^{2}-1$
(3) $\mathrm{g}(\mathrm{x})=3 \mathrm{x}^{2}-2, \mathrm{x}=\left(\frac{2}{\sqrt{3}}, \frac{-2}{\sqrt{3}}\right)$
Sol:
We know that
$g(x)=3 x^{2}-2$
Given that, $x=\left(\frac{2}{\sqrt{3}}, \frac{-2}{\sqrt{3}}\right)$
Substitute $x=\frac{2}{\sqrt{3}}$ in $g(x)$
$g\left(\frac{2}{\sqrt{3}}\right)=3\left(\frac{2}{\sqrt{3}}\right)^{2}-2$
$=3\left(\frac{4}{3}\right)-2$
$=4-2$
$=2 \neq 0$
Now, Substitute $x=\frac{-2}{\sqrt{3}}$ in $g(x)$
$g\left(\frac{-2}{\sqrt{3}}\right)=3\left(\frac{-2}{\sqrt{3}}\right)^{2}-2$
$=3\left(\frac{4}{3}\right)-2$
$=4-2$
$=2 \neq 0$
Since, the results when $x=\left(\frac{2}{\sqrt{3}}, \frac{-2}{\sqrt{3}}\right)$ are not 0, they are roots of $3 x^{2}-2$
(4) $p(x)=x^{3}-6 x^{2}+11 x-6, x=1,2,3$
Sol:
We know that,
$p(x)=x^{3}-6 x^{2}+11 x-6$
given that the values of $x$ are $1,2,3$
substitute $x=1$ in $p(x)$
$p(1)=1^{3}-6(1)^{2}+11(1)-6$
$=1-(6 \star 1)+11-6$
$=1-6+11-6$
$=0$
Now, substitute $x=2$ in $p(x)$
$P(2)=2^{3}-6(2)^{2}+11(2)-6$
$=(2 * 3)-(6 * 4)+(11 * 2)-6$
$=8-24-22-6$
$=0$
Now, substitute $x=3$ in $p(x)$
$P(3)=3^{3}-6(3)^{2}+11(3)-6$
$=(3 * 3)-(6 * 9)+(11 * 3)-6$
$=27-54+33-6$
$=0$
Since, the result is 0 for $x=1,2,3$ these are the roots $o f x^{3}-6 x^{2}+11 x-6$
(5) $\mathrm{f}(\mathrm{x})=5 \mathrm{x}-\pi, \mathrm{x}=\frac{4}{5}$
we know that,
$f(x)=5 x-\pi$
Given that, $x=\frac{4}{5}$
Substitute the value of $x$ in $f(x)$
$f\left(\frac{4}{5}\right)=5\left(\frac{4}{5}\right)-\pi$
$=4-\pi$
$\neq 0$
Since, the result is not equal to zero, $x=\frac{4}{5}$ is not the root of the polynomial $5 x-\pi$
(6) $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}, \mathrm{x}=0$
Sol:
we know that, $f(x)=x^{2}$
Given that value of $x$ is ‘ 0 ‘
Substitute the value of $x$ in $f(x)$
$f(0)=0^{2}$
$=0$
Since, the result is zero, $x=0$ is the root of $x^{2}$
(7) $\mathrm{f}(\mathrm{x})=\mathrm{lx}+\mathrm{m}, \mathrm{x}=\frac{-\mathrm{m}}{\mathrm{l}}$
Sol:
We know that,
$f(x)=\mid x+m$
Given, that $x=\frac{-m}{1}$
Substitute the value of $x$ in $f(x)$
$f\left(\frac{-m}{1}\right)=1\left(\frac{-m}{1}\right)+m$
$=-m+m$
$=0$
Since, the result is $0, x=\frac{-m}{1}$ is the root of $I x+m$
(8) $\mathrm{f}(\mathrm{x})=2 \mathrm{x}+1, \mathrm{x}=\frac{1}{2}$
Sol:
We know that,
$f(x)=2 x+1$
Given that $\mathrm{x}=\frac{1}{2}$
Substitute the value of $x$ and $f(x)$
$f\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)+1$
$=1+1$
$=2 \neq 0$
Since, the result is not equal to zero
$x=\frac{1}{2}$ is the root of $2 x+1$

Q3. If $x=2$ is a root of the polynomial $f(x)=2 x^{2}-3 x+7 a$, Find the value of $a$
Sol:
We know that, $f(x)=2 x^{2}-3 x+7 a$
Given that $x=2$ is the root of $f(x)$
Substitute the value of $x$ in $f(x)$
$f(2)=2(2)^{2}-3(2)+7 a$
$=(2 * 4)-6+7 a$
$=8-6+7 a$
$=7 a+2$
Now, equate $7 a+2$ to zero
$\Rightarrow 7 a+2=0$
$\Rightarrow 7 a=-2$
$\Rightarrow \mathrm{a}=\frac{-2}{7}$
The value of $\mathrm{a}=\frac{-2}{7}$

Q4. If $x=\frac{-1}{2}$ is zero of the polynomial $p(x)=8 x^{3}-a x^{2}-x+2$, Find the value of a
Sol:
We know that, $p(x)=8 x^{3}-a x^{2}-x+2$
Given that the value of $x=\frac{-1}{2}$
Substitute the value of $x$ in $f(x)$
$p\left(\frac{-1}{2}\right)=8\left(\frac{-1}{2}\right)^{3}-a\left(\frac{-1}{2}\right)^{2}-\left(\frac{-1}{2}\right)+2$
$=-8\left(\frac{1}{8}\right)-a\left(\frac{1}{4}\right)+\frac{1}{2}+2$
$=-1-\left(\frac{a}{4}+\frac{1}{2}+2\right.$
$=1-\left(\frac{a}{4}+\frac{1}{2}\right.$
$=\frac{3}{2}-\frac{a}{4}$
To, find the value of a , equate $\mathrm{p}\left(\frac{-1}{2}\right)$ to zero
$p\left(\frac{-1}{2}\right)=0$
$\frac{3}{2}-\frac{a}{4}=0$
On taking L.C.M
$\frac{6-a}{4}=0$
$\Rightarrow 6-a=0$
$\Rightarrow \quad a=6$

Q5. If $x=0$ and $x=-1$ are the roots of the polynomial $f(x)=2 x^{3}-3 x^{2}+a x+b$, Find the of $a$ and $b$.
Sol:
We know that, $f(x)=2 x^{3}-3 x^{2}+a x+b$
Given , the values of $x$ are 0 and $-1$
Substitute $x=0$ in $f(x)$
$f(0)=2(0)^{3}-3(0)^{2}+a(0)+b$
$=0-0+0+b$
$=b \quad—1$
Substitute $x=(-1)$ in $f(x)$
$f(-1)=2(-1)^{3}-3(-1)^{2}+a(-1)+b$
$=-2-3-a+b$
$=-5-a+b \quad—–2$
We need to equate equations 1 and 2 to zero
$b=0$ and $-5-a+b=0$
since, the value of $b$ is zero
substitute $\mathrm{b}=0$ in equation 2
$\Rightarrow-5-a=-b$
$\Rightarrow-5-a=0$
$a=-5$
the values of $\mathrm{a}$ and $\mathrm{b}$ are $-5$ and 0 respectively

Q6. Find the integral roots of the polynomial $f(x)=x^{3}+6 x^{2}+11 x+6$
Sol:
Given, that $f(x)=x^{3}+6 x^{2}+11 x+6$
Clearly we can say that, the polynomial $\mathrm{f}(\mathrm{x})$ with an integer coefficient and the highest degree term coefficient which is known as leading factor is 1 .
So, the roots of $f(x)$ are limited to integer factor of 6 , they are
$\pm 1, \pm 2, \pm 3, \pm 6$
Let $x=-1$
$f(-1)=(-1)^{3}+6(-1)^{2}+11(-1)+6$
$=-1+6-11+6$
$=0$
Let $x=-2$
$f(-2)=(-2)^{3}+6(-2)^{2}+11(-2)+6$
$=-8-(6 * 4)-22+6$
$=-8+24-22+6$
$=0$
Let $x=-3$
$f(-3)=(-3)^{3}+6(-3)^{2}+11(-3)+6$
$=-27-(6 \star 9)-33+6$
$=-27+54-33+6$
$=0$
But from all the given factors only $-1,-2,-3$ gives the result as zero.
So, the integral multiples of $\mathrm{x}^{3}+6 \mathrm{x}^{2}+11 \mathrm{x}+6$ are $-1,-2,-3$

Q7. Find the rational roots of the polynomial $f(x)=2 x^{3}+x^{2}-7 x-6$
Sol:
Given that $f(x)=2 x^{3}+x^{2}-7 x-6$
$f(x)$ is a cubic polynomial with an integer coefficient . If the rational root in the form of $\frac{p}{q}$, the values of $p$ are
limited to factors of 6 which are $\pm 1, \pm 2, \pm 3, \pm 6$
and the values of $q$ are limited to the highest degree coefficient i.e 2 which are $\pm 1, \pm 2$
here, the possible rational roots are
$\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$
Let, $x=-1$
$f(-1)=2(-1)^{3}+(-1)^{2}-7(-1)-6$
$=-2+1+7-6$
$=-8+8$
$=0$
Let, $x=2$
$f(-2)=2(2)^{3}+(2)^{2}-7(2)-6$
$=(2 * 8)+4-14-6$
$=16+4-14-6$
$=20-20$
$=0$
Let, $x=\frac{-3}{2}$
$f\left(\frac{-3}{2}\right)=2\left(\frac{-3}{2}\right)^{3}+\left(\frac{-3}{2}\right)^{2}-7\left(\frac{-3}{2}\right)-6$
$=2\left(\frac{-27}{8}\right)+\frac{9}{4}-7\left(\frac{-3}{2}\right)-6$
$=\left(\frac{-27}{4}\right)+\frac{9}{4}-\left(\frac{-21}{2}\right)-6$
$=-6.75+2.25+10.5-6$
$=12.75-12.75$
$=0$
But from all the factors only $-1,2$ and $\frac{-3}{2}$ gives the result as zero
So, the rational roots of $2 \mathrm{x}^{3}+\mathrm{x}^{2}-7 \mathrm{x}-6$ are $-1,2$ and $\frac{-3}{2}$

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