Question:
Find the condition that the zeros of the polynomial f(x) = x3 + 3px2 + 3qx + r may be in A.P.
Solution:
Let $a-d, a$ and $a+d$ be the zeros of the polynomials $f(x)$. Then,
Sum of the zeros $=\frac{\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$
$a-d+a+a+d=\frac{-3 p}{1}$
$3 a=-3 p$
$a=\frac{-3 \times p}{3}$
$a=-p$
Since $a$ is a zero of the polynomial $f(x)$. Therefore,
$f(x)=x^{3}+3 p x^{2}+3 q x+r$
$f(a)=0$
$f(a)=a^{3}+3 p a^{2}+3 q a+r$
$a^{3}+3 p a^{2}+3 q a+r=0$
Substituting $a=-p$ we get,
$(-p)^{3}+3 p(-p)^{2}+3 q(-p)+r=0$
$-p^{3}+3 p^{3}-3 p q+r=0$
$2 p^{3}-3 p q+r=0$
Hence, the condition for the given polynomial is $2 p^{3}-3 p q+r=0$.