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 \[{{x}^{3}}=1\]
\[\begin{align}
  & 3a=\dfrac{-3p}{1} \\
 & \therefore a=-p \\
\end{align}\]
Since ‘a’ is a zero of the polynomial f(x),
\[\begin{align}
  & \therefore f(x)={{x}^{3}}+3p{{x}^{2}}+3qx+r \\
 & f(a)=0 \\
\end{align}\]
Put \[x=a.\]
\[\begin{align}
  & f(a)={{a}^{3}}+3p{{a}^{2}}+3qa+r \\
 & f(a)=0 \\
 & \Rightarrow {{a}^{3}}+3p{{a}^{2}}+3qa+r=0 \\
\end{align}\]
Substitute \[a=-p.\]
\[\begin{align}
  & \therefore {{(-p)}^{3}}+3p{{(-p)}^{2}}+3q(-p)+r=0 \\
 & -{{p}^{3}}+3{{p}^{3}}+3pq+r=0 \\
 & \Rightarrow 2{{p}^{3}}-3pq+r=0 \\
\end{align}\]
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.