Question

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

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

Answer 1

Let p(y) = ${\displaystyle 7y^2-\frac{11}{3}y-\frac{2}{3}}$

= 21y² – 11y – 2

= 21y² – 14y + 3y – 2

= (3y – 2)(7y + 2)

So, the zeroes of p(y) are ${\displaystyle \frac{2}{3}}$ and ${\displaystyle -\frac{1}{7}}$

∴ Sum of zeroes = ${\displaystyle \frac{2}{3}-\frac{1}{7}}$

= ${\displaystyle \frac{11}{21}}$

= ${\displaystyle -\left(\frac{-11}{3\times 7}\right)}$

= ${\displaystyle \frac{-(\text{coefficient of} y)}{\text{coefficient of} y^2}}$

And product of zeroes = ${\displaystyle \left(\frac{2}{3}\right)(-\frac{1}{7})}$

= ${\displaystyle \frac{-2}{3\times 7}}$

= ${\displaystyle \frac{\text{constant term}}{\text{coefficient of} y^2}}$