# How do you factor by grouping 8r^3 - 64r^2 + r -8?

Apr 25, 2018

$\left(8 {r}^{2} + 1\right) \left(r - 8\right)$

#### Explanation:

In grouping, separate the first two terms (in this case, $8 {r}^{3}$ and $- 64 {r}^{2}$) from the second two terms (in this case, $r$ and $- 8$):

$\left(8 {r}^{3} - 64 {r}^{2}\right) + \left(r - 8\right) \rightarrow$ Factor out the GCF from each of the pairs in the parentheses

$8 {r}^{2} \left(r - 8\right) + 1 \left(r - 8\right) \rightarrow$ Combine the $8 {r}^{2}$ and $1$

$\left(8 {r}^{2} + 1\right) \left(r - 8\right)$