Question:
If the zeros of the polynomial $1(x)=x^{3}-12 x^{2}+39 x+k$ 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 $=\frac{\text { Coefficient of } x^{2}}{\text { Coefficient of } x^{3}}$
$a-d+a+a+d=\frac{-(-12)}{1}$
$a-\boldsymbol{A}+a+a+\boldsymbol{A}=12$
$3 a=12$
$a=\frac{12}{3}$
$a=4$
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 $-28$.
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