If $x^{3} + 8 x^{2} + k x + 18$ is completely divisible by $x^{2} + 6 x + 9$ , then find the value of 'k'.

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Solution

$x^{3} + 8 x^{2} + k x + 18$ is completely divisible by $x^{2} + 6 x + 9$

First of ll we should break $x^{2} = 6 x + 9$

$\Rightarrow x^{2} + 2.3 x + x^{2} + (x + 3)^{2}$

Hence $x = - 3$ is the zero of $x^{3} + 8 x^{2} + k x + 18$

Now put $x = - 3$ in $x^{3} + 8 x + k x + 18$

$\Rightarrow$ $(- 3)^{3} + 8 (- 3)^{2} + k (- 3) + 18 = 0$

$\Rightarrow$ $- 27 + 8 \times 9 - 3 k + 18 = 0$

$\Rightarrow$ $- 27 + 72 - 3 k + 18 = 0$

$\Rightarrow$ $63 - 3 k = 0$

$\Rightarrow$ $k = 63 / 3$

$\Rightarrow$ $k = 21$

$\Rightarrow$ $k = 21$

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