Factor x^3 + 3x^2 - 4x - 12 completely.

Step-by-step solution

1. 1

   Choose grouping for a four-term cubic. There is no factor common to all four terms, so split them into two pairs and hope each pair leaves the same binomial. That is exactly what happens here.

2. 2

   Group and factor each pair.

   $\left(x^3+3x^2\right)+\left(-4x-12\right)=x^2(x+3)-4(x+3)$

   Pulling out $-4$ (not $+4$) from the second pair is what makes the binomials match; taking $+4$ would leave $(-x-3)$ instead.

3. 3

   Factor out the common binomial.

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

4. 4

   Finish with the difference of squares. Since $x^2-4=x^2-2^2$:

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

   so

   $x^3+3x^2-4x-12=(x+3)(x-2)(x+2)$

   Stopping at $(x+3)(x^2-4)$ is the most common way to lose marks on this problem.

5. 5

   Read off the roots. The expression is zero at $x=-3,\ 2,\ -2$ — three distinct real roots, all accounted for by a cubic.

6. 6

   Verify by expansion. Comparing the original with the triple product at every integer from $-30$ to $29$ gives exact agreement ✓. Spot check at $x=1$: $1+3-4-12=-12$, and $(4)(-1)(3)=-12$ ✓.