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I've started reading through a pre-calculus textbook for self-study and came across this problem in the second chapter: $$(x-2)^3-(x-2)^2$$ The final answer is $(x-2)^2(x-3)$

Everywhere I look but I have no idea how it comes to that. Binomial expansion was mentioned on one site but the binomial theorem appears in a later chapter. What is the method/methods used to factor this expression?

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    $\begingroup$ Let $\ y = x\!-\!2.\ $ Then it is $\ y^3 - y^2 = y^2(y - 1) = (x\!-\!2)^2 (x\!-\!2 - 1)\ \ $ $\endgroup$ – Bill Dubuque Jun 25 '15 at 22:47
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Use the distributive property: $$ab-ac=a(b-c)$$ Here, we have $$a=(x-2)^2\qquad b=(x-2)\qquad c=1$$ Thus $$(x-2)^3-(x-2)^2=(x-2)^2((x-2) -1)=(x-2)^2(x-3)$$

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  • $\begingroup$ Thank you! Now it makes sense. $\endgroup$ – dantastic Jun 25 '15 at 23:06

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