The value of $p$ for which the polynomial $2 x^{4} + 3 x^{3} + 2 p x^{2} + 3 x + 6$ is exactly divisible by $(x + 2)$ is

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Solution

The correct option is C $- 1$
Let $p (x)$ = $2 x^{4} + 3 x^{3} + 2 p x^{2} + 3 x + 6$
By factor theorem, $p (x)$ will be exactly divisible by $(x + 2)$ if $p ( -2) = 0. $N o w,$ p(-2) = 2( -2)^4 + 3( -2)^3 + 2p( -2)^2 + 3( -2) + 6= 32 - 24 + 8p - 6 + 6= 8 + 8p$
Since, $p (- 2) = 0$
$\Rightarrow 8 + 8 p = 0$
$\Rightarrow 8 p = 0$
$\Rightarrow 8 p = - 8$
$\Rightarrow p = - 1$

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