Question

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

$7y^2-\frac{11}{3}y-\frac{2}{3}$.

• A. $\frac{11}{3},-\frac{1}{7}$
• B. $\frac{7}{3},-\frac{1}{7}$
• C. $\frac{1}{3},-\frac{1}{7}$
• D. $\frac{2}{3},-\frac{1}{7}$

Answer 1

The correct option is D $\frac{2}{3},-\frac{1}{7}$

The quadratic polynomial is $7y^2-\frac{11}{3}y-\frac{2}{3}$.
The zeros of the polynomial are:
$7y^2-\frac{11}{3}y-\frac{2}{3}=0$.

$⟹$ $7y^2-\frac{14}{3}y+y-\frac{2}{3}=0$

$⟹$ $7y(y-\frac{2}{3})+(y-\frac{2}{3})=0$

$⟹$ $(y-\frac{2}{3})(7y+1)=0$

$⟹$ $y=\frac{2}{3}$ and $y=-\frac{1}{7}$.

The zeros are $\frac{2}{3}$ and $-\frac{1}{7}$.

Now verifying the relation of zeros with the coefficients of the quadratic polynomial $7y^2-\frac{11}{3}y-\frac{2}{3}$:
Comparing it with the standard from of quadratic equation $a{x}^2+bx+c$, we get, $a=7$, $b=-\frac{11}{3}$ and $c=\frac{2}{3}$.
Thus,
Sum of the zeros $=-\frac{b}{a}$

$⟹$ $\frac{2}{3}+(-\frac{1}{7})$ $=\frac{2}{3}-\frac{1}{7}$

$=\frac{14-3}{21}$ $=\frac{11}{3\times 7}$ $=-\frac{b}{a}$.

Product of the zeros $=\frac{c}{a}$

$⟹$ $=\frac{2}{3}(-\frac{1}{7})$ $=-\frac{2}{21}$ $=\frac{c}{a}$.

Therefore, hence verified.

Hence, option $D$ is correct.