factoring of $ e^{2x}-3e^{x}+2 = 0 $

Question

How does

$ e^{2x}-3e^{x}+2 = 0 $

factors to

$ (e^x - 1) (e^x - 2 ) $

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Because when I try to factor:

$ e^{2x}-3e^{x}+2 = 0 $

$ e^{2x}-2e^{x}-1e^{x}+2 = 0 $

But $-2e^{x} * -1e^{x}$ should give 2 e^x^2 instead of $2e^{2x}$

Answer 1

Your mistakenly think that

$$e^x \cdot e^x=e^{x^2}$$

But $e^a \cdot e^b =e^{a+b}$. Your confusion comes most likely from confusing $\left( e^x \right)^2$ which is actually $e^{2x}$ with $e^{x^2}$. Anyhow

$$\left( e^x \right)^2 \neq e^{x^2}$$

Answer 2

put $t=e^{x}$. You get

$$t^2 -3t + 2$$

Factor this one (by finding its roots,for example), and substitute back $t=e^{x}$

Answer 3

Hint:

use $t=e^x$ and note that for the second degree trinomial that come from we have $1 \cdot 2=2$ and $-(1+2)=-3$.