Question: How do I compute $-6(-4)^{n-1}$ + $8(-4)^{n-2}$ ?

I recall that as long as the number from both operands (in this case: -4) are the same, I can actually "add" them together. But the problem is the -6 from the first operand and the +8 from the second operand makes the math difficult right?

So should I just factor it by $(-4)^{n-2}$? 

What are my options?

Answer: 
$$(-4)^{n-1}(-6+8(-4)^{-1})$$
$$(-4)^{n-1}(-6-2)$$
$$(-4)^{n}\frac{-8}{-4}=2(-4)^n$$
Hints: $-6(-4)^{n-1}=-6\times (-4)\times (-4)^{n-2}=24(-4)^{n-2}$