Find the $x$ intercepts of $x^3+x^2-4x-4$ I am to find the $x$ intercepts of $x^3+x^2-4x-4$.
The solution in my book shows these as: $(2,0)$, $(-2,0)$, $(-1,0)$ whereas I get just $(4, 0)$, $(-1, 0)$.
My working:
$x^3+x^2-4x-4$ =
$x^2(x+1)-4(x+1)$ =
$(x^2-4)(x+1)$
Thus, intercepts when $x$ is equal to $4$ and $-1$.
Where have I gone wrong and how can I arrive at $x$ intercepts of $2, -2$ and $1$?

Just as said in the comments, I will quickly put this into answer form.
The $x^2-4$ can be simplified to $(x+2)(x-2)$, so the roots are $2$, $-2$, and $-1$.
