Question

Find the condition that the zeros of the polynomial $f(x)={{x}^{3}}+3p{{x}^{2}}+3qx+r$ are in an A.P.

Answer 1

Hint: Consider $a-d,a,a+d$ as a sum of polynomials. Find the expression for the sum of zeroes. Put $f(a)=0$ and find the condition by substituting $a=-p$ which is obtained from the sum of zeroes.

“Complete step-by-step answer:”
Given the polynomial,$f(x)={{x}^{3}}+3p{{c}^{2}}+3qx+r$
Let $a-d,a,a+d$ be the zeroes of the polynomial f(x) which is in A.P. with common difference d.
The sum of zeroes$=\dfrac{-Coefficient of{{x}^{2}}}{Coefficientof{{x}^{3}}}=(a-d)+a+(a+d)$
$\Rightarrow (a-d)+a+(a+d)=\dfrac{-3p}{1}$
where coefficient of ${{x}^{2}}=3p.$
Coefficient of ${{x}^{3}}=1$

& 3a=\dfrac{-3p}{1} \\\ & \therefore a=-p \\\ \end{aligned}$$ Since ‘a’ is a zero of the polynomial f(x), $$\begin{aligned} & \therefore f(x)={{x}^{3}}+3p{{x}^{2}}+3qx+r \\\ & f(a)=0 \\\ \end{aligned}$$ Put $$x=a.$$ $$\begin{aligned} & f(a)={{a}^{3}}+3p{{a}^{2}}+3qa+r \\\ & f(a)=0 \\\ & \Rightarrow {{a}^{3}}+3p{{a}^{2}}+3qa+r=0 \\\ \end{aligned}$$ Substitute $$a=-p.$$ $$\begin{aligned} & \therefore {{(-p)}^{3}}+3p{{(-p)}^{2}}+3q(-p)+r=0 \\\ & -{{p}^{3}}+3{{p}^{3}}+3pq+r=0 \\\ & \Rightarrow 2{{p}^{3}}-3pq+r=0 \\\ \end{aligned}$$ Hence the condition for the given polynomial is $$2{{p}^{3}}-3pq+r=0$$. Note: We might think that for taking conditions, you might want to do differentiation, but that’s not right. Taking f’(x) won’t give us the required condition. As we have found $$a=-p$$, find$$f(a)=0$$, and then substitute $$a=-p$$ to get the required condition.