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

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Solution

Let roots be $a - d, a, a + d$ (As in A.P)
$f (x) = x^{3} - 12 x^{2} + 39 x + k$
Sum= $a - d + a + a + d = 12$
$3 a = 12$
$a = 4$
Sum taken 2 at a time= $(a - d) a + a (a + d) + (a - d) (a + d)$
$39 = 3 a^{2} - d^{2}$
$d^{2} = 48 - 39$
$d^{2} = 9$
$d = 3$
$∴$ Roots are $4, 7, 1$
Product $= k = - 4.7.1$
$\Rightarrow - 28$

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