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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.

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    $\begingroup$ Are you just asking why $ab - ac = a(b - c)$? $\endgroup$
    – user296602
    Mar 1, 2018 at 17:06
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    $\begingroup$ Did you try a few values of $n$? $\endgroup$ Mar 1, 2018 at 17:08
  • $\begingroup$ Welcome to MSE. Please read this text about how to ask a good question. $\endgroup$ Mar 1, 2018 at 17:09

3 Answers 3

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

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Note that

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

Following your way

$$...=2^{-3}\cdot \:2^n-2^{-2}\cdot \:2^n=2^n(2^{-3}-2^{-2})=2^n2^{-3}(1-2^1)=-2^{n-3}$$

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It is more intuitive for you turn them into fractions:

$$2^{n-3}-2^{n-2}=\frac{2^n}{2^3}-\frac{2^n}{2^2}=2^n\Biggl(\frac1{2^3}-\frac1{2^2}\Biggr)=2^n\Biggl(\frac{2^2-2^3}{2^5}\Biggr)=$$ $$2^n\Biggl(-\frac{2^2}{2^5}\Biggr)=2^n\Biggl(-\frac{1}{2^3}\Biggr)=-2^n(2^{-3})=-2^{n-3}$$

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    $\begingroup$ Note that a shortcut may be $$2^n\left(\frac1{2^3}-\frac1{2^2}\right)=2^n\left(\frac1{2^3}-\frac2{2^3}\right)=2^n\left(-\frac1{2^3}\right)=-2^{n-3}$$ $\endgroup$
    – TheSimpliFire
    Mar 1, 2018 at 19:13
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    $\begingroup$ Haven't really thought about it that way. Thanks for the answer. $\endgroup$
    – Skatinima
    Mar 2, 2018 at 11:55
  • $\begingroup$ @TheSimpliFire True that. It's only shown for the sake of discussion. $\endgroup$
    – John Glenn
    Mar 2, 2018 at 13:42

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