Question

Find the zeros of the following polynomial by factorization method and verify the relations between the zeroes and the coefficient of the polynomials :
$v^2+4\sqrt{3}v-15$

Answer 1

Let $f\left(v\right)=v^2+4v\sqrt{3}-15$

For zeroes of $f\left(v\right),f\left(v\right)=0$

$\Rightarrow v^2+4\sqrt{3}v-15=0$

$\Rightarrow v^2+5\sqrt{3}v-1\sqrt{3}v-15=0$ $\left[\begin{array}{c}15=5\times 3\\ =1\times 5\times \sqrt{3}\times \sqrt{3}\end{array}\right]$

$\Rightarrow v(v+5\sqrt{3})-\sqrt{3}(v+5\sqrt{3})=0$

$\Rightarrow (v+5\sqrt{3})(v-\sqrt{3})=0$

$\Rightarrow (v+5\sqrt{3})=0$ or $(v-\sqrt{3})=0$

$\Rightarrow v=-5\sqrt{3}$ or $v=\sqrt{3}$

verification:

$\alpha =-5\sqrt{3},\beta =\sqrt{3}$

$a=1,b=4\sqrt{3}$, and $c=-15$

$\alpha +\beta =\frac{-b}{a}$

$\Rightarrow -5\sqrt{3}+\sqrt{3}=\frac{-4\sqrt{3}}{1}$

$\Rightarrow -4\sqrt{3}=-4\sqrt{3}$

$\Rightarrow LHS=RHS$

Hence, verified.

$\alpha .\beta =\frac{c}{a}$

$\Rightarrow (-5\sqrt{3})\left(\sqrt{3}\right)=\frac{-15}{1}$

$\Rightarrow -5\times 3=-15$

$\to -15=-15$

$\Rightarrow LHS=RHS$

Hence, verified.