Factor 2x^2 - 10x - 12 completely.

Step-by-step solution

1. 1

   Check for a common numerical factor before anything else. All three coefficients $2$, $-10$, $-12$ are divisible by $2$:

   $2x^2-10x-12=2\left(x^2-5x-6\right)$

   Doing this first is much easier than running the AC method on the original, whose $ac$ would be $-24$.

2. 2

   Factor the monic bracket. Two numbers with product $-6$ and sum $-5$:

   $(-6)\cdot 1=-6,\qquad -6+1=-5$

   so $x^2-5x-6=(x-6)(x+1)$.

3. 3

   Assemble the complete factorisation. The constant $2$ must be kept — dropping it changes the expression:

   $2x^2-10x-12=2(x-6)(x+1)$

4. 4

   Verify by expanding. $(x-6)(x+1)=x^2-5x-6$, and doubling gives $2x^2-10x-12$ $\checkmark$. At $x=0$: original $=-12$, and $2(-6)(1)=-12$ $\checkmark$. The roots of the corresponding equation are $x=6$ and $x=-1$; a constant factor never affects the roots, only the vertical scale.