If the zeros of the polynomial $f(x)\ =\ ax^3\ +\ 3bx^2\ +\ 3cx\ +\ d$ are in A.P., prove that $2b^3\ -\ 3abc\ +\ a^2d\ =\ 0$.

Given:

The zeros of the polynomial $f(x)\ =\ ax^3\ +\ 3bx^2\ +\ 3cx\ +\ d$ are in A.P.


To do:

Here, we have to prove that $2b^3\ -\ 3abc\ +\ a^2d\ =\ 0$.


Solution:

 Let the zeros of the given polynomial be α, β and γ.


Given that the zeros are in A.P.

So, consider the roots as, $α = p – d$, $β = p$ and $γ = p +d$ where, $p$ is the first term and $d$ is the common difference.

Comparing $f(x) $ with the standard form of a cubic polynomial,  $a= a$, $b= 3b$, $c= 3c$ and $d=d$

Therefore, 

Sum of the roots $= α + β + γ = (p– d) + p + (p + d) = 3p = \frac{-b}{a} =\frac{-3b}{a} = -\frac{3b}{a}$

$3p= -\frac{3b}{a}$

$p= -\frac{b}{a}$

$β = p$ is a root of the given polynomial.

This implies,

$f(p)=0$

$f(-\frac{b}{a})=a(-\frac{b}{a})^3+3b(-\frac{b}{a})^2+3c(-\frac{b}{a})+d=0$

$ \begin{array}{l}
a\left(\frac{-b^{3}}{a^{3}}\right) +3b\left(\frac{b^{2}}{a^{2}}\right) +3c\left(\frac{-b}{a}\right) +d=0\\
\\
\frac{-b^{3}}{a^{2}} +\frac{3b^{3}}{a^{2}} -\frac{3bc}{a} +d=0\\
\\
\frac{-b^{3} +3b^{3} -3abc+a^{2} d}{a^{2}} =0\\
\\
2b^{3} -3abc+a^{2} d=0
\end{array}$

Hence proved.

Updated on: 10-Oct-2022

22 Views

Related Articles

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements