How do you factor by grouping $n^3-2n^2+4n-8$ ?

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Answer 1 · 2018-03-27T00:33:34

$(n-2)(n^2+4)$

$n^3-2n^2+4n-8$

Factor $n^2$ from the first two terms and factor $4$ from the last 2 terms.

$n^2(n-2)+4(n-2)$

You now have two terms both containing a factor of $(n-2)$ . Factor $(n-2)$ from both of these terms.

$(n-2)(n^2+4)$

Hey, that was easy!

How do you factor by grouping n^3-2n^2+4n-8n^3-2n^2+4n-8?