Exercise 3.3 - Chapter 3 Algebra 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.3$
Question $1 .$
Check whether $\mathrm{p}(\mathrm{x})$ is a multiple of $\mathrm{g}(\mathrm{x})$ or not.
$\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-5 \mathrm{x}^{2}+4 \mathrm{x}-3, \mathrm{~g}(\mathrm{x})=\mathrm{x}-2$
Solution:
$\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-5 \mathrm{x}^{2}+4 \mathrm{x}-3 ; \mathrm{g}(\mathrm{x})=\mathrm{x}-2$ Let $\mathrm{g}(\mathrm{x})=0$ $\mathrm{x}-2=0$ $\mathrm{x}=2$ $\mathrm{p}(2)=2^{3}-5\left(2^{2}\right)+4(2)-3$ $=8-5 \times 4+8-3=8-20+5=-7 \neq 0$ $\Rightarrow \mathrm{p}(\mathrm{x})$ is not a multiple of $\mathrm{g}(\mathrm{x})$
 

Question $2 .$
By remainder theorem, find the remainder when, $\mathrm{p}(\mathrm{x})$ is divided by $\mathrm{g}(\mathrm{x})$ where,
(i) $\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-2 \mathrm{x}^{2}-4 \mathrm{x}-1 ; \mathrm{g}(\mathrm{x})=\mathrm{x}+1$
(ii) $\mathrm{p}(\mathrm{x})=4 \mathrm{x}^{3}-12 \mathrm{x}^{2}+14 \mathrm{x}-3 ; \mathrm{g}(\mathrm{x})=2 \mathrm{x}-1$
(iii) $\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-3 \mathrm{x}^{2}+4 \mathrm{x}+50 ; \mathrm{g}(\mathrm{x})=\mathrm{x}-3$
Solution:
(i) $\mathrm{p}(\mathrm{x})=\mathrm{x}^{3}-2 \mathrm{x}^{2}-4 \mathrm{x}-1 ; \mathrm{g}(\mathrm{x})=\mathrm{x}+1$
Let $g(x)=x+1$
$\mathrm{x}+\mathrm{l}=0$
$\mathrm{X}=-1$
$\mathrm{P}(-1)=(-1)^{3}-2(-1)^{2}-4(-1)-1$
$\begin{aligned}
&=-1-2 \times 1+4-1 \\
&=-4+4=0 \\
&\therefore \text { Remainder }=0 .
\end{aligned}$

(ii) $\mathrm{p}(\mathrm{x})=4 \mathrm{x}^{3}-12 \mathrm{x}^{2}+14 \mathrm{x}-3 ; \mathrm{g}(\mathrm{x})=2 \mathrm{x}-1$
$\begin{array}{rlrl}p\left(\frac{1}{2}\right) & =4\left(\frac{1}{2}\right)^{3}-12\left(\frac{1}{2}\right)^{2}+14\left(\frac{1}{2}\right)-3 & \\ & =4 \times \frac{1}{8}-12 \times \frac{1}{4}+7-3 & \text { Let } g(x)=2 x-1 \\ & =\frac{1}{2}-3+4 & 2 x-1 & =0 \\ & =\frac{1-6+8}{2}=\frac{3}{2} & 2 x=1 \\ \therefore \text { Remainder } & =\frac{3}{2} .\end{array}$
$=\frac{1-6+8}{2}=\frac{3}{2}$
$\therefore$ Remainder $=\frac{3}{2}$.
(iii) $p(x)=x^{3}-3 x^{2}+4 x+50 ; g(x)=x-3$
Let $g(x)=x-3$
$x-3=0$
$x=3$
$\mathrm{p}(3)=3^{3}-3\left(3^{2}\right)+4(3)+50$
$=27-27+12+50$
$=62$
$\therefore$ Remainder $=62$.
 

Question $3 .$
Find the remainder when ${ }^{3}-4 x^{2}+7 x-5$ is divided by $(x+3)$
Solution:
$\left(3 x^{3}-4 x^{2}+7 x-5\right)+(x+3)$

Question $4 .$
What is the remainder when $x^{2018}+2018$ is divided by $x-1$.
Solution:
$\mathrm{x}^{2018}+2018$ is divided by $\mathrm{x}-1$
Let $g(x)=x-1=0$
$x=1$
$\begin{aligned}
&p(x)=x^{2018}+2018 \\
&p(1)=1^{2018}+2018 \\
&=1+2018=2019
\end{aligned}$

Question 5
For what value of $k$ is the polynomial $p(x)=2 x^{3}-k x^{2}+3 x+10$ exactly divisible by $(x-2)$.
Solution:
Let $g(x)=x-2=0$
$x=2$
Since $p(x)$ is exactly divisible by $(x-2)$
$\begin{aligned}
&\mathrm{p}(2)=2\left(2^{3}\right)-\mathrm{k}\left(2^{2}\right)+3(2)+10 \\
&=16-4 \mathrm{k}+6+10 \\
&=32-4 \mathrm{k}=0 \\
&=-\mathrm{k}=-32 \\
&\mathrm{k}=\frac{32}{4}=8 .
\end{aligned}$

Question $6 .$
If two polynomials $2 x^{3}+a x^{2}+4 x-12$ and $x^{3}+x^{2}-2 x+$ a leave the same remainder when divided by $(x-3)$, find the value of $a$. and also find the remainder.
Solution:
Let $f(x)=2 x^{3}+a x^{2}+4 x-12$ and $g(x)=x^{3}+x^{2}-2 x+a$
When $f(x)$ is divided by $x-3$, the remainder is $f(3)$.
Now $\mathrm{f}(3)=2(3)^{3}+\mathrm{a}(3)^{2}+4(3)-12=54+9 \mathrm{a}+12-12$
$\mathrm{f}(3)=9 \mathrm{a}+54 \ldots \ldots \ldots \ldots(1)$
When $g(x)$ is divided by $x-3$, the remainder is $g(3)$.
Now $\mathrm{g}(3)=3^{3}+3^{2}-2(3)+\mathrm{a}=27+9-6+\mathrm{a}$
$\mathrm{g}(3)=\mathrm{a}+30 \ldots \ldots \ldots .(2)$
Since, the remainder's are same $(1)=(2)$
Given that $\mathrm{f}(3)=\mathrm{g}(3)$
That is $9 a+54=a+30$
$9 \mathrm{a}-\mathrm{a}=30-54 \Rightarrow 8 \mathrm{a}=-24 \therefore \mathrm{a}=-3$
Substituting $a=-3$ in $f(3)$, we get
$f(3)=9(-3)+54=-27+54$
$f(3)=27$
$\therefore$ The remainder is 27 .
 

Question $7 .$
Determine whether $(x-1)$ is a factor of the following polynomials:
(i) $x^{3}+5 x^{2}-10 x+4$
(ii) $x^{4}+5 x^{2}-5 x+1$
Solution:
(i) Let $\mathrm{P}(\mathrm{x})=\mathrm{x}^{3}+5 \mathrm{x}^{2}-10 \mathrm{x}+4$
By factor theorem $(x-1)$ is a factor of $P(x)$, if $P(1)=0$

$P(1)=1^{3}+5\left(1^{2}\right)-10(1)+4=1+5-10+4$ $P(1)=0$ $\therefore(x-1)$ is a factor of $x^{3}+5 x^{2}-10 x+4$
(ii) Let $\mathrm{P}(\mathrm{x})=\mathrm{x}^{4}+5 \mathrm{x}^{2}-5 \mathrm{x}+1$
By/actor theorem, $(\mathrm{x}-1)$ is a factor of $\mathrm{P}(\mathrm{x})$, if $\mathrm{P}(1)=0$
$\mathrm{P}(1)=1^{4}+5\left(1^{2}\right)-5(1)+1=1+5-5+1=2 \neq 0$
$\therefore(x-1)$ is not a factor of $x^{4}+5 x^{2}-5 x+1$
 

Question $8 .$
Using factor theorem, show that $(x-5)$ is a factor of the polynomial $2 x^{3}-5 x^{2}-28 x+15$
Solution:
$\operatorname{Let} P(x)=2 x^{3}-5 x^{2}-28 x+15$
By factor theorem, $(x-5)$ is a factor of $P(x)$, if $P(5)=0$
$P(5)=2(5)^{2}-5(5)^{2}-28(5)+15$ $=2 \times 125-5 \times 25-140+15$ $=250-125-140+15=265-265=0$ $\therefore(x-5)$ is a factor of $2 x^{3}-5 x^{2}-28 x+15$
 

Question 9
. Determine the value of $m$, if $(x+3)$ is a factor of $x^{3}-3 x^{2}-m x+24$
Solution:
Let $P(x)=x^{3}-3 x^{2}-m x+24$
By using factor theorem,
$(x+3)$ is a factor of $P(x)$, then $P(-3)=0$
$\begin{aligned}
&P(-3)=(-3)^{3}-3(-3)^{2}-m(-3)+24=0 \\
&\Rightarrow-27-3 \times 9+3 m+24=0 \Rightarrow 3 m=54-24 \\
&\Rightarrow m=\frac{30}{3}=10
\end{aligned}$

Question $10 .$
If both $(x-2)$ and $\left(x-\frac{1}{2}\right)$ are the factors of $a x^{2}+5 x+b$, then show that $a=b$.
Solution:
Let $P(x)=a x^{2}+5 x+b$
$(x-2)$ is a factor of $P(x)$, if $P(2)=0$
$\begin{aligned}
&\mathrm{P}(2)=\mathrm{a}(2)^{2}+5(2)+\mathrm{b}=0 \\
&4 \mathrm{a}+10+\mathrm{b}=0 \\
&4 \mathrm{a}+\mathrm{b}=-10 \ldots \ldots \ldots \ldots \ldots(\mathrm{l} . \ldots \ldots \mathrm{l}) \\
&\left(\mathrm{x}-\frac{1}{2}\right) \text { is a factor of } \mathrm{P}(\mathrm{x}), \mathrm{P}\left(\frac{1}{2}\right)=0
\end{aligned}$
$\begin{aligned}
\mathrm{P}\left(\frac{1}{2}\right) &=a\left(\frac{1}{2}\right)^{2}+5\left(\frac{1}{2}\right)+b=0 \\
\frac{a}{4}+\frac{5}{2}+b &=0
\end{aligned}$

$\begin{aligned}
&\frac{a}{4}+b=\frac{-5}{2} \\
&\frac{a+4 b}{4}=\frac{-5}{2} \\
&2 \mathrm{a}+8 \mathrm{~b}=-20 \\
&\mathrm{a}+4 \mathrm{~b}=-10 \ldots \ldots \ldots \ldots \ldots \ldots(2) \\
&\text { From }(1) \text { and }(2) \\
&4 \mathrm{a}+\mathrm{b}=-10 \ldots \ldots \ldots \ldots(1) \\
&\mathrm{a}+4 \mathrm{~b}=-10 \ldots \ldots \ldots \ldots(2) \\
&(1) \text { and }(2) \Rightarrow 4 \mathrm{a}+\mathrm{b}=\mathrm{a}+4 \mathrm{~b} \\
&3 \mathrm{a}=3 \mathrm{~b} \\
&\therefore \mathrm{a}=\mathrm{b} . \text { Hence it is proved. }
\end{aligned}$

Question $11 .$
If $(x-1)$ divides the polynomial $k x^{3}-2 x^{2}+25 x-26$ without remainder, then find the value of $k$. Solution:
Let $\mathrm{P}(\mathrm{x})=\mathrm{kx}^{3}-2 \mathrm{x}^{2}+25 \mathrm{x}-26$
By factor theorem, $(x-1)$ divides $\mathrm{P}(\mathrm{x})$ without remainder, $\mathrm{P}(1)=0$
$\begin{aligned}
&\mathrm{P}(1)=\mathrm{k}(1)^{3}-2(1)^{2}+25(1)-26=0 \\
&\mathrm{k}-2+25-26=0 \\
&\mathrm{k}-3=0 \\
&\mathrm{k}=3
\end{aligned}$

Question $12 .$
Check if $(x+2)$ and $(x-4)$ are the sides of a rectangle whose area is $x^{2}-2 x-8$ by using factor theorem.
Solution:
Let $P(x)=x^{2}-2 x^{2}-8$
By using factor theorem, $(x+2)$ is a factor of $P(x)$, if $P(-2)=0$
$P(-2)=(-2)^{2}-2(-2)-8=4+4-8=0$
and also $(x-4)$ is a factor of $P(x)$, if $P(4)=0$
$\mathrm{p}(4)=4^{2}-2(4)-8=16-8-8=0$
$\therefore(\mathrm{x}+2),(\mathrm{x}-4)$ are the sides of a rectangle whose area is $\mathrm{x}^{2}-2 \mathrm{x}-8$.