Find the condition that the zeros of the polynomial  $f(x)\ =\ x^3\ +\ 3px^2\ +\ 3qx\ +\ r$  may be in A.P.

Given:

The zeros of the polynomial $f(x)\ =\ x^3\ +\ 3px^2\ +\ 3qx\ +\ r$ are in A.P.


To do:

Here, we have to find the condition such that the zeros of the given polynomial are in A.P. 


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, 

$α = s - d$, $β = s$ and $γ = s +d$ where, $s$ is the first term and $d$ is the common difference.

Comparing $f(x) $ with the standard form of a cubic polynomial, 

$a= 1$, $b= 3p$, $c= 3q$ and $d=r$

Therefore, 

Sum of the roots $= α + β + γ = (s– d) +s + (s + d) = 3s = \frac{-b}{a} =\frac{-3p}{1} = -3p$

$3s= -3p$

$s=-p$

$β = s$ is a root of the polynomial $f(x)$.

This implies,

$f(s)=0$

$f(-p)=(-p)^3+3p(-p)^2+3q(-p)+r=0$

$-p^3+3p^3-3pq+r=0$

$2p^3-3pq+r=0$

The condition that the zeros of the polynomial  $f(x)\ =\ x^3\ +\ 3px^2\ +\ 3qx\ +\ r$  may be in A.P. is $2p^3-3pq+r=0$.

Updated on: 10-Oct-2022

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