If the zeros of the polynomial $f(x)\ =\ x^3\ -\ 12x^2\ +\ 39x\ +\ k$ are in A.P., find the value of $k$.


Given:

The zeros of the polynomial $f(x)\ =\ x^3\ -\ 12x^2\ +\ 39x\ +\ k$ are in A.P.


To do:

Here, we have to find the value of $k$. 


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, 
$a= 1$, $b= -12$, $c= 39$ and $d=k$

Therefore, 

Sum of the roots $= α + β + γ = (p– d) + p + (p + d) = 3p = \frac{-b}{a} =\frac{-(-12)}{1} = 12$

$3p= 12$

$p= \frac{12}{3}=4$

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

This implies,

$f(p)=0$

$f(4)=(4)^3-12(4)^2+39(4)+k=0$

$64-12(16)+156+k=0$

$64-192+156+k=0$

$220-192+k=0$

$28+k=0$

$k=-28$

The value of $k$ is $-28$.

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Updated on: 10-Oct-2022

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