Question

Find the zeroes of the following polynomials by factorisation method and verify the relation between the zeroes and the coefficients of the polynomials:

$4x^2+5\sqrt{2}x-3$

Answer 1

$4x^2+5\sqrt{2}x-3$
$=4x^2+6\sqrt{2}x-\sqrt{2}x-3$
$=2\sqrt{2}x(\sqrt{2}x+3)-(\sqrt{2}x+3)$
$=(\sqrt{2}x+3)(2\sqrt{2}x-1)$
So, the zeros of the polynomial are
$(\sqrt{2}x+3)=0$ and $(2\sqrt{2}x-1)=0$
$x=\frac{-3}{\sqrt{2}}$ and $x=\frac{1}{2\sqrt{2}}$
$x=\frac{-3\sqrt{2}}{2}$ and $x=\frac{\sqrt{2}}{4}$
Now, the polynomial is $4x^2+5\sqrt{2}x-3$
Comparing it with $a{x}^2+bx+c=0$, we get $a=4$, $b=5\sqrt{2}$ and $c=-3$
Sum of the zeros $=\frac{-3\sqrt{2}}{2}+\frac{\sqrt{2}}{4}$
$=\frac{-6\sqrt{2}+\sqrt{2}}{4}$
$=\frac{-5\sqrt{2}}{4}$
$=\frac{-b}{a}$
Product of the zeros $=\left(\frac{-3\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{4}\right)$
$=\left(\frac{-3}{2}\right)\left(\frac{2}{4}\right)$
$=\frac{-3}{4}$
$=\frac{c}{a}$