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If $α$ and $β$ are the zeros of the polynomial $f(x)\ =\ x^2\ +\ px\ +\ q$, form a polynomial whose zeros are $(α\ +\ β)^2$ and $(α\ -\ β)^2$.
Given:
$α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ +\ px\ +\ q$.
To do:
We have to find the quadratic polynomial having $(α\ +\ β)^2$ and $(α\ -\ β)^2$ as its zeros.
Solution:
We know that,
The standard form of a quadratic equation is $ax^2+bx+c=0$, where a, b and c are
constants and $a≠0$
Comparing the given equation with the standard form of a quadratic equation,
$a=1$, $b=p$ and $c=q$
Sum of the roots $= α+β = \frac{-b}{a} = \frac{-p}{1}=-p$.
Product of the roots $= αβ = \frac{c}{a} = \frac{q}{1}=q$.
Let the sum and product of the zeros of the given quadratic equation be $S$ and $P$.
Therefore,
$ \begin{array}{l}
S=( \alpha +\beta )^{2} +( \alpha -\beta )^{2}\\
\\
\ \ =\alpha ^{2} +\beta ^{2} +2\alpha \beta +\alpha ^{2} +\beta ^{2} -2\alpha \beta \\
\\
\ \ =2\left( \alpha ^{2} +\beta ^{2}\right)\\
\\
\ \ =2\left(( \alpha +\beta )^{2} -2\alpha \beta \right)\\
\\
\ \ =2\left(( -p)^{2} -2( q)\right)\\
\\
\ \ =2\left( p^{2} -2q\right)\\
\\
P=( \alpha +\beta )^{2} \times ( \alpha -\beta )^{2}\\
\\
\ \ \ =\left( \alpha ^{2} +\beta ^{2} +2\alpha \beta \right)\left( \alpha ^{2} +\beta ^{2} -2\alpha \beta \right)\\
\\
\ \ \ =\left(( \alpha +\beta )^{2} -2\alpha \beta +2\alpha \beta \right)\left(( \alpha +\beta )^{2} -2\alpha \beta -2\alpha \beta \right)\\
\\
\ \ \ =( \alpha +\beta )^{2}\left(( \alpha +\beta )^{2} -4\alpha \beta \right)\\
\\
\ \ \ =p^{2}\left( p^{2} -4( -q)\right)\\
\\
\ \ \ =p^{2}\left( p^{2} -4q\right)
\end{array}$
The quadratic polynomial having the sum of the roots $S$ and product of the roots $P$ is $f(x)=k(x^2-(S)x+P)$, where $k$ is any non-zero real number.
Therefore,
$f(x)=k(x^2-(2(p^2-q))x+(p^2(p^2-4q)))$
$f(x)=k(x^2-2(p^2-q)x+p^2(p^2-4q))$
The required quadratic polynomial is $f(x)=k(x^2-2(p^2-q)x+p^2(p^2-4q))$, where $k$ is any non-zero real number.
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