Question

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

$v^2+4\sqrt{3}v-15$.

• A. $\sqrt{3},+5\sqrt{3}$
• B. $\sqrt{3},-5\sqrt{3}$
• C. $\sqrt{3},-5$
• D. $\sqrt{3},-5\sqrt{2}$

Answer 1

The correct option is B $\sqrt{3},-5\sqrt{3}$

Here, $v^2+4\sqrt{3}v-15$
$=v^2+5\sqrt{3}v-\sqrt{3}v-15$
$=v(v+5\sqrt{3})-\sqrt{3}(v+5\sqrt{3})$
$=(v+5\sqrt{3})(v-\sqrt{3})$.
So, the zeros of the polynomial are :
$(v+5\sqrt{3})=0$ and $(v-\sqrt{3})=0$
$v=-5\sqrt{3}$ and $v=\sqrt{3}$.
Now, the polynomial is $v^2+4\sqrt{3}v-15$.

To verify the relation between the zeros and the coefficients of the polynomial:
Compare it with $a{x}^2+bx+c=0$, we get $a=1$, $b=4\sqrt{3}$ and $c=-15$.
Sum of the zeros $=-5\sqrt{3}+\sqrt{3}$
$=-4\sqrt{3}$
$=\frac{-b}{a}$.
Product of the zeros $=(-5\sqrt{3})\left(\sqrt{3}\right)$
$=(-5\sqrt{3})\left(\sqrt{3}\right)$
$=-5\left(3\right)$
$=-15$
$=\frac{c}{a}$.

Hence, verified.

Therefore, option $B$ is correct.