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Question

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


Solution

$$x^3+ 8x^2 + kx +18$$ is completely divisible by $$x^2 +6x+9$$

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

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

Hence $$x=-3$$ is the zero of $$x^3+8x^2+kx+18$$

Now put $$x=-3$$ in $$x^3+8x+kx+18$$

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

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

$$\Rightarrow$$ $$ -27 +72-3k+18=0$$

$$\Rightarrow$$ $$63-3k=0$$

$$\Rightarrow$$ $$k=63/3$$

$$\Rightarrow$$ $$k=21$$

$$\Rightarrow$$ $$k=21$$

Mathematics

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