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

Solution:

If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}+p x+q$

$\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$=\frac{-p}{1}$

$\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$=\frac{q}{1}$

= q

Let $S$ and $P$ denote respectively the sums and product of the zeros of the polynomial whose zeros are $(\alpha+\beta)^{2}$ and $(\alpha-\beta)^{2}$. Then,

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

$S=2\left[\alpha^{2}+\beta^{2}\right]$

$S=2\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]$

$S=2\left(p^{2}-2 \times q\right)$

$S=2\left(p^{2}-2 q\right)$

$S=2\left(p^{2}-2 q\right)$

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

$P=\left(\alpha^{2}+\beta^{2}+2 \alpha \beta\right)\left(\alpha^{2}+\beta^{2}-2 \alpha \beta\right)$

$P=(p)^{2}\left((p)^{2}-4 \times q\right)$

$P=p^{2}\left(p^{2}-4 q\right)$

The required polynomial of $f(x)=k\left(k x^{2}-s x+p\right)$ is given by

$f(x)=k\left\{x^{2}-2\left(p^{2}-2 q\right) x+p^{2}\left(p^{2}-4 q\right)\right\}$

$f(x)=k\left\{x^{2}-2\left(p^{2}-2 q\right) x+p^{2}\left(p^{2}-4 q\right)\right\}$, where $k$ is any non-zero real number.