Exercise 2.6-Additional Questions - Chapter 2 - Algebra - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions
Question 1.
Find the zeros of the polynomial function $f(x)=9 x^2-36$ Solution:
$
9 x^2-36=(3 x)^2-6^2
$

If $x=2$ is one root of $x^3+2 x^2-5 x-6=0$ then find the other roots of the equation Solution: $\mathrm{x}=2$ is a root so applying synthetic division

$\therefore$ The other factor is $\mathrm{x}^2+4 \mathrm{x}+3$
Now $x^3+2 x^2-5 \mathrm{x}-6=(\mathrm{x}-2)\left(\mathrm{x}^2+4 \mathrm{x}+3\right)$
$\therefore \mathrm{x}^3+2 \mathrm{x}^2-5 \mathrm{x}-6=0 \Rightarrow(\mathrm{x}-2)\left(\mathrm{x}^2+4 \mathrm{x}+3\right)$
$\mathrm{x}-2=0$ or $\mathrm{x}^2+4 \mathrm{x}+3=0$
$\mathrm{x}=2$ or $(\mathrm{x}+1)(\mathrm{x}+3)=0$
$\Rightarrow \mathrm{x}=-1$ or -3
so the roots are $\mathrm{x}=-1,2,-3$