# If the zeros of the polynomial f(x) = x3 − 12x2 + 39x + k are in A.P., find the value of k.

Question:

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

Solution:

Let $a-d, a$ and $a+d$ be the zeros of the polynomial $f(x)$

Then,

Sum of the zeros $=\frac{\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$

$a-d+a+a+d=\frac{-(-12)}{1}$

$3 a=12$

$a=\frac{12}{3}$

$a=4$

Since $a$ is a zero of the polynomial $f(x)$

$f(x)=x-12 x+39 x+k$

$f(a)=0$

$f(a)=4^{3}-12 \times 4^{2}+39 \times 4+k$

$0=64-192+156+k$

$0=220-192+k$

$0=28+k$

$-28=k$

Hence, the value of $k$ is $-28$.