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$x^{3}+x^{2}-4 x-4$

Let $f(x)=x^{3}+x^{2}-4 x-4$ be the given polynomial. The factors of the constant term $-4$ are $±1,±2,±4$

We have,

$f(-1)=(-1)^{3}+(-1)^{2}-4(-1)-4=-1+1+4-4=0$

And, $f(2)=(2)^{3}+(2)^{2}-4(2)-4=8+4-8-4=0$

So, $\quad(x+1)$ and $(x-2)$ are factors of $f(x)$

$\Rightarrow \quad(x+1)(x-2)$ is also a factor of $f(x)$

$\Rightarrow \quad x^{2}-x-2$ is a factor of $f(x)$

Let us know divide $f(x)=x^{3}+x^{2}-4 x-4$ by $x^{2}-x-2$ to get the other factors of $f(x).$

By long division, we have

$\begin{array}{l}x^{2}-x-2 |\overline {x^{3}+x^{2}-4 x-4} (x+2)\\ \;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\; x^{3}-x^{2}+2 x\;\;\;\;\;\;\; \\ \hline \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2 x^{2}-2 x-4 \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2 x^{2}-2 x-4 \\ \hline \;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\; 0 \end{array}$

$\therefore x^{3}+x^{2}-4 x-4=\left(x^{2}-x-2\right)(x+2)$

$\Rightarrow x^{3}+x^{2}-4 x-4=(x+1)(x-2)(x+2)$

Hence, $x^{3}+x^{2}-4 x-4=(x-2)(x+1)(x+2)$