# Factoring $2^{^{n-3}}-2^{^{n-2}}$ (not homework question)

At the moment, I'm studying for my math exam, and I came upon a problem which involves factoring the powers of this polynomial:

$2^{^{n-3}}-2^{^{n-2}}$

After a few minutes of being stuck, i looked up a solution and found this (from symbolab):

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$=2^{-3}\cdot \:2^n-2^{-2}\cdot \:2^n$

$\mathrm{Factor\:out\:common\:term\:}2^{-3}\cdot \:2^n$

$=2^{-3}\left(1-2\right)\cdot \:2^n$

$\mathrm{Refine}$

$=-2^{n-3}$

My question is:

how can i factor

$2^{-3}\cdot \:2^n$

from

$2^{-3}\cdot \:2^n-2^{-2}\cdot \:2^n$

to get

$2^{-3}\left(1-2\right)\cdot \:2^n$ ?

I can't seem to understand the logic behind this factorization.

If you could help me understand it, I would really appreciate it.

• Are you just asking why $ab - ac = a(b - c)$? – user296602 Mar 1 '18 at 17:06
• Did you try a few values of $n$? – Angina Seng Mar 1 '18 at 17:08
• Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos Mar 1 '18 at 17:09

$$2^{n-3}-2^{n-2}=2^{(n-3)+0}-2^{(n-3)+1}=2^{n-3}(2^0-2^1)=\boxed{-2^{n-3}}$$

Note that

$$\large{2^{^{n-3}}-2^{^{n-2}}=2^{^{n-3}}(1-2^1)=-2^{n-3}}$$

$$...=2^{-3}\cdot \:2^n-2^{-2}\cdot \:2^n=2^n(2^{-3}-2^{-2})=2^n2^{-3}(1-2^1)=-2^{n-3}$$
$$2^{n-3}-2^{n-2}=\frac{2^n}{2^3}-\frac{2^n}{2^2}=2^n\Biggl(\frac1{2^3}-\frac1{2^2}\Biggr)=2^n\Biggl(\frac{2^2-2^3}{2^5}\Biggr)=$$ $$2^n\Biggl(-\frac{2^2}{2^5}\Biggr)=2^n\Biggl(-\frac{1}{2^3}\Biggr)=-2^n(2^{-3})=-2^{n-3}$$
• Note that a shortcut may be $$2^n\left(\frac1{2^3}-\frac1{2^2}\right)=2^n\left(\frac1{2^3}-\frac2{2^3}\right)=2^n\left(-\frac1{2^3}\right)=-2^{n-3}$$ – TheSimpliFire Mar 1 '18 at 19:13