Question

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

(x) $7y^2–\frac{11}{3}y-\frac{2}{3}$

Answer 1

x) Let $f\left(y\right)=7y^2–\frac{11}{3}y-\frac{2}{3}$
$=21y^2–11y–2$
$=21y^2–14y+3y–2$ [by splitting the middle term]
$=7y(3y–2)+1(3y–2)$
$=(3y–2)(7y+1)$
So, the value of $7y^2–\frac{11}{3}y–\frac{2}{3}$ is zero when 3y – 2 = 0 or 7y + 1 = 0
i.e., when $y=\frac{2}{3}ory=-\frac{1}{7}$
so, the zeroes $7y^2–\frac{11}{3}y-\frac{2}{3}are\frac{2}{3}and-\frac{1}{7}$
$\therefore$ sum of zeroes =$\frac{2}{3}-\frac{1}{7}=\frac{14-3}{21}=\frac{11}{21}=-\left(\frac{-11}{3\times 7}\right)$
$=(-1)\left(\frac{coefficientofy}{coefficientof{y}^2}\right)$
And product of zeroes =$\left(\frac{2}{3}\right)(-\frac{1}{7})=(-\frac{2}{21})$
$=\frac{Constantterm}{Coefficientof{y}^2}$
Hence, verified the relations between the zeroes and the coefficients of the polynomial.