Exercise 3.8 - Chapter 3 Algebra 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.8$
Question 1.
Factorise each of the following polynomials using synthetic division:
(i) $\mathrm{x}^{3}-3 \mathrm{x}^{2}-10 \mathrm{x}+24$
(ii) $2 x^{3}-3 x^{2}-3 x+2$
(iii) $-7 \mathrm{x}+3+4 \mathrm{x}^{3}$
(iv) $x^{3}+x^{2}-14 x-24$
(v) $x^{3}-7 x+6$
(vi) $x^{3}-10 x^{2}-x+10$
Solution:
(i) $x^{3}-3 x^{2}-10 x+24$
Let $p(x)=x^{3}-3 x^{2}-10 x+24$
Sum of all the co-efficients $=1-3-10+24=25-13=12 \neq 0$
Hence $(x-1)$ is not a factor.
Sum of co-efficient of even powers with constant $=-3+24=21$
Sum of co-efficients of odd powers $=1-10=-9$
$21 \neq-9$
Hence $(x+1)$ is not a factor.
$\begin{aligned}
&\mathrm{p}(2)=2^{3}-3\left(2^{2}\right)-10 \times 2+24=8-12-20+24 \\
&=32-32=0 \therefore(x-2) \text { is a factor. }
\end{aligned}$
Now we use synthetic division to find other factor

Thqs $(x-2)(x+3)(x-4)$ are the factors.
$\therefore \mathrm{x}^{3}-3 \mathrm{x}^{2}-10 \mathrm{x}+24=(\mathrm{x}-2)(\mathrm{x}+3)(\mathrm{x}-4)$
(ii) $2 x^{2}-3 x^{2}-3 x+2$
Let $p(x)=2 x^{3}-3 x^{2}-3 x+2$
Sum of all the co-efficients are
$2-3-3+2=4-6=-2 \neq 0$
$\therefore(\mathrm{x}-1)$ is not a factor
Sum of co-efficients of even powers of $x$ with constant $=-3+2=-1$
Sum of co-efficients of odd powers of $x=2-3=-1$
$(-1)=(-1)$
$\therefore(\mathrm{x}+1)$ is a factor
Let us find the other factors using synthetic division

Quotient is $2 x^{2}-5 x+2=2 x-4 x-x+2=2 x(x-2)-1(x-2)$ $=(x-2)(2 x-1)$
$\therefore 2 \mathrm{x}^{3}-3 \mathrm{x}^{2}-3 \mathrm{x}+2=(\mathrm{x}+1)(\mathrm{x}-2)(2 \mathrm{x}-1)$
(iii) $-7 \mathrm{x}+3+4 \mathrm{x}^{3}$
Let $p(x)=4 x^{3}+0 x^{2}-7 x+3$
Sum of the co-efficients are $=4+0-7+3$
$=7-7=0$
$\therefore(\mathrm{x}-1)$ is a factor
Sum of co-efficients of even powers of $x$ with constant $=0+3=3$
Sum of co-efficients of odd powers of $x$ with constant $=4-7=-3$
$-3 \neq-3$
$\therefore(\mathrm{x}+1)$ is not a factor
Using synthetic division, let us find the other factors.

Quotient is $4 x^{2}+4 x-3$
$=4 x^{2}+6 x-2 x-3$ $=2 x(2 x+3)-1(2 x+3)$ $=(2 x+3)(2 x-1)$ $\therefore$ The factors are $(x-1),(2 x+3)$ and $(2 x-1)$ $\therefore-7 x+3+4 x^{3}=(x+1)(2 x+3)(2 x-1)$
$\therefore-7 \mathrm{x}+3+4 \mathrm{x}^{3}=(\mathrm{x}+1)(2 \mathrm{x}+3)(2 \mathrm{x}-1)$
(iv) $x^{3}+x^{2}-14 x-24$
Let $p(x)=x^{3}+x^{2}-14 x-24$
Sum of the co-efficients are $=1+1-14-24=-36 \neq 0$
$\therefore(x-1)$ is not a factor
Sum of co-efficients of even powers of $x$ with constant $=1-24=-23$
Sum of co-efficients of odd powers of $x=1-14=-3$
$-23 \neq-13$
$\therefore(\mathrm{x}+1)$ is also not a factor
$\mathrm{p}(2)=2^{3}+2^{2}-14(2)-24=8+4-28-24$
$=12-52 \neq 0$, $(x-2)$ is a not a factor
$p(-2)=(-2)^{3}+(-2)^{2}-14(-2)-24$
$=-8+4+28-24=32-32=0$
$\therefore(\mathrm{x}+2)$ is a factor
To find the other factors let us use synthetic division.

$\therefore$ The factors are $(x+2),(x+3),(x+4)$
$\therefore \mathrm{x}^{3}+\mathrm{x}^{2}-14 \mathrm{x}-24=(\mathrm{x}+2)(\mathrm{x}+3)(\mathrm{x}-4)$
(v) $x^{3}-7 x+6$
Let $p(x)=x^{3}+0 x^{2}-7 x+6$
Sum of the co-efficients are $=1+0-7+6=7-7=0$ $\therefore(\mathrm{x}-1)$ is a factor
Sum of co-efficients of even powers of $x$ with constant $=0+6=6$
Sum of coefficient of odd powers of $x=1-7=-7$
$6 \neq-7$
$\therefore(\mathrm{x}+1)$ is not a factor
To find the other factors, let us use synthetic division.

$\therefore$ The factors are $(\mathrm{x}-1),(\mathrm{x}-2),(\mathrm{x}+3)$
$\therefore \mathrm{x}^{3}+0 \mathrm{x}^{2}-7 \mathrm{x}+6=(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}+3)$
(vi) $x^{3}-10 x^{2}-x+10$
Let $p(x)=x^{3}-10 x^{2}-x+10$
Sum of the co-efficients $=1-0-1+10$
$=11-11=0$
$\therefore(\mathrm{x}-1)$ is a factor
Sum of co-efficients of even powers of $x$ with constant $=-10+10=0$
Sum of co-efficients of odd powers of $=1-1=0$
$\therefore(\mathrm{x}+1)$ is a factor
Synthetic division