If α and β are the zeros of the quadratic polynomial p(x)

Question:

If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $p(x)=4 x^{2}-5 x-1$, find the value of $\alpha^{2} \beta+\alpha \beta^{2}$.

Solution:

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomials $p(x)=4 x^{2}-5 x-1$

Sum of the zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$\alpha+\beta=-\left(-\frac{5}{4}\right)$

$\alpha+\beta=\frac{5}{4}$

Product of zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$\alpha \beta=-\frac{1}{4}$

We have, $\alpha^{2} \beta+\alpha \beta^{2}$

$\alpha^{2} \beta+\alpha \beta^{2}=\alpha \beta(\alpha+\beta)$

By substituting $\alpha+\beta=\frac{5}{4}$ and $\alpha \beta=-\frac{1}{4}$ in $\alpha^{2} \beta+\alpha \beta^{2}=\alpha \beta(\alpha+\beta)$, we get

$\alpha^{2} \beta+\alpha \beta^{2}=-\frac{1}{4}\left(\frac{5}{4}\right)$

$\alpha^{2} \beta+\alpha \beta^{2}=-\frac{1}{4} \times \frac{5}{4}$

$\alpha^{2} \beta+\alpha \beta^{2}=-\frac{5}{16}$

Hence, the value of $\alpha^{2} \beta+\alpha \beta^{2}$ is $-\frac{5}{16}$.