Factor x^3 + x^2 - 4x - 4 completely.

Step-by-step solution

1. 1

   Choose grouping, and see why it is the right tool. With exactly four terms and no common factor across all of them, grouping in pairs is the standard first attempt. It works whenever the two pairs leave the same binomial behind — which is what we check next.

2. 2

   Split into two pairs.

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

   Keeping the signs attached to the terms as you group prevents the classic sign error in the second pair.

3. 3

   Factor each pair. From the first, pull out $x^2$; from the second, pull out $-4$ (not $+4$, so that the binomial matches):

   $x^3+x^2=x^2(x+1),\qquad -4x-4=-4(x+1)$

   Both pairs leave $(x+1)$ — the grouping has worked.

4. 4

   Factor out the common binomial.

   $x^2(x+1)-4(x+1)=(x+1)\left(x^2-4\right)$

5. 5

   Finish with the difference of squares. $x^2-4=x^2-2^2$, so

   $x^2-4=(x-2)(x+2)$

   and therefore

   $x^3+x^2-4x-4=(x+1)(x-2)(x+2)$

   Stopping at $(x+1)(x^2-4)$ would leave the factoring incomplete.

6. 6

   Verify by expansion over many values. Comparing the original cubic with the product at every integer from $-40$ to $39$ gives exact agreement ✓. The roots $-1,\ 2,\ -2$ are also easy to confirm directly: $(-1)^3+(-1)^2+4-4=0$, $8+4-8-4=0$, $-8+4+8-4=0$ ✓.