Question

Question 1 (viii)
Find the zeroes of the following polynomials by factorization method and verify the relations between the zeroes and the coefficient of the polynomials

(viii) $v^2+4\sqrt{3}v-15$

Answer 1

(viii) Let $f\left(v\right)=v^2+4\sqrt{3}v-15$
$=v^2+(5\sqrt{3}-\sqrt{3})v–15$ [by splitting the middle term]
$=v^2+5\sqrt{3}v-\sqrt{3}v-15$
$=v(v+5\sqrt{3})-\sqrt{3}(v+5\sqrt{3})$
$=(v+5\sqrt{3})(v-\sqrt{3})$
So, the value of $v^2+4\sqrt{3}v–15$ is zero when $v+5\sqrt{3}=0orv-\sqrt{3}=0$
i.e., when $v=-5\sqrt{3}orv=\sqrt{3}$
So, the Sum of zeroes of is $-4\sqrt{3}$
= $(-1)\left(\frac{coefficientofv}{coefficientof{v}^2}\right)$
And product of zeroes = $(-5\sqrt{3})\left(\sqrt{3}\right)$
$=-5\times 3=-15$
$=(-1{)}^2\left(\frac{constantterm}{coefficientof{v}^2}\right)$
Hence, verified the relations between the zeroes and the coefficients of the polynomial.