If $\alpha$ and $\beta$ are the zeros of the polynomial $1(x)=x^{2}+p x+q$, from a polynomial whose zeros are $(\alpha+\beta)^{2}$ and $(\alpha-\beta)^{2}$.
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.