Find the quadratic polynomial whose zeros are $\frac{2}{3}$ and $\frac{-1}{4}$. Verify the relation between the coefficients and the zeros of the polynomial.
Let $\alpha=\frac{2}{3}$ and $\beta=\frac{-1}{4}$.
Sum of the zeroes $=(\alpha+\beta)=\frac{2}{3}+\left(\frac{-1}{4}\right)=\frac{8-3}{12}=\frac{5}{12}$
Product of the zeroes $=\alpha \beta=\frac{2}{3} \times\left(\frac{-1}{4}\right)=\frac{-\frac{2^{1}}{12^{6}}}{12^{2}}=\frac{-1}{6}$
$\therefore$ Required polynomial $=x^{2}-(\alpha+\beta) x+\alpha \beta=x^{2}-\frac{5}{12} x+\left(\frac{-1}{6}\right)$
$=x^{2}-\frac{5}{12} x-\frac{1}{6}$
Sum of the zeroes $=\frac{5}{12}=\frac{-(\text { coefficient of } x)}{\left(\text { coefficient of } x^{2}\right)}$
Product of the zeroes $=\frac{-1}{6}=\frac{\text { constant term }}{\text { coefficient of } x^{2}}$