Use the factor theorem to factorize completely: $\;$ $x^3 + x^2 -4x - 4$
Let $\;$ $f \left(x\right) = x^3 + x^2 - 4 x - 4$
$f \left(2\right) = 2^3 + 2^2 -4 \times 2 - 4 = 0$
$\implies$ $\left(x - 2\right)$ is a factor of $f \left(x\right)$ $\;\;\;$ [By factor theorem]
Now,
$\begin{array}{lllll}
x - 2 & ) & x^3 + x^2 - 4x - 4 & ( & x^2 + 3x + 2 \\
& & x^3 - 2x^2 & & \\
& & - - - - - - - - - - - - & & \\
& & \ \ \ \ \ \ \ \ 3x^2 - 4x & & \\
& & \ \ \ \ \ \ \ \ 3x^2 - 6x & & \\
& & - - - - - - - - - - - - & & \\
& & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x - 4 & & \\
& & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x - 4 & & \\
& & - - - - - - - - - - - - & & \\
& & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0
\end{array}$
$\begin{aligned}
\therefore \; x^3 + x^2 - 4x - 4 & = \left(x - 2\right) \left(x^2 + 3x + 2\right) \\\\
& = \left(x - 2\right) \left(x^2 + 2 x + x + 2\right) \\\\
& = \left(x - 2\right) \left[x \left(x + 2\right) + 1 \left(x + 2\right)\right] \\\\
& = \left(x - 2\right) \left(x + 1\right) \left(x + 2\right)
\end{aligned}$