Find $k$ so that $(x^{2} + 2 x + k)$ is a factor of $(2 x^{4} + x^{3} + - 14 x^{2} + 5 x + 6)$ .

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Solution

$2 x^{2} - 3 x - (8 + 2 k)$
$x^{2} + 2 x + k) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 2 x^{4} + x^{3} - 14 x^{2} + 5 x + 6$

$2 x^{2} + 4 x^{3} + 2 k x^{2}$
$- - -$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 3 x^{3} - (14 + 2 k) x^{2} + 5 x$

$- 3 x^{2} - 6 x^{2} - 3 k x$
$+ + +$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - (8 + 2 k) x^{2} + (5 + 3 k) x + 6$

$- (8 + 2 k) x^{2} - (16 + 4 k) x - (8 k + 2 k^{2})$
$+ + +$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ (21 + 7 k) x + 2 k^{2} + 8 k + 6 = 0$

Comparing the coefficient of $x$ we get
$21 + 7 k = 0$
$\Rightarrow$ $7 k = - 21$
$∴$ $k = - 3$

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