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

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 1

$$(-4)^{n-1}(-6+8(-4)^{-1})$$ $$(-4)^{n-1}(-6-2)$$ $$(-4)^{n}\frac{-8}{-4}=2(-4)^n$$

Answer 2

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

Answer 3

This problem is looking to assess your comfort with the properties of exponents. Recall these general properties...

• $x^a x^b = x^{a+b}$
• $\dfrac{x^a}{x^b} = x^{a-b}$
• $(x^a)^b = (x^b)^a = x^{ab}$
• $x^{-a} = \dfrac{1}{x^a}$ for $x \ne 0$

So, solving the problem using a slightly alternative method than the others (for the sake of variety):

$$-6(-4)^{n-1} + 8(-4)^{n-2}$$ $$ = -6(-4)^{-1}(-4)^n + 8(-4)^{-2}(-4)^n$$ $$ = \dfrac{-6}{-4}(-4)^n + \dfrac{8}{4^2}(-4)^n$$ $$ = (-4)^n\left(\dfrac{3}{2} + \dfrac{1}{2}\right)$$ $$ = 2(-4)^n$$