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

Mar 27, 2018

$\left(n - 2\right) \left({n}^{2} + 4\right)$

Explanation:

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

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

${n}^{2} \left(n - 2\right) + 4 \left(n - 2\right)$

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

$\left(n - 2\right) \left({n}^{2} + 4\right)$

Hey, that was easy!