# How do you multiply (4v+8)(8v^2-3v-4)?

Mar 17, 2017

See the entire solution process below:

#### Explanation:

To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

$\left(\textcolor{red}{4 v} + \textcolor{red}{8}\right) \left(\textcolor{b l u e}{8 {v}^{2}} - \textcolor{b l u e}{3 v} - \textcolor{b l u e}{4}\right)$ becomes:

$\left(\textcolor{red}{4 v} \times \textcolor{b l u e}{8 {v}^{2}}\right) - \left(\textcolor{red}{4 v} \times \textcolor{b l u e}{3 v}\right) - \left(\textcolor{red}{4 v} \times \textcolor{b l u e}{4}\right) + \left(\textcolor{red}{8} \times \textcolor{b l u e}{8 {v}^{2}}\right) - \left(\textcolor{red}{8} \times \textcolor{b l u e}{3 v}\right) - \left(\textcolor{red}{8} \times \textcolor{b l u e}{4}\right)$

$32 {v}^{3} - 12 {v}^{2} - 16 v + 64 {v}^{2} - 24 v - 32$

We can now group and then combine like terms:

$32 {v}^{3} + 64 {v}^{2} - 12 {v}^{2} - 16 v - 24 v - 32$

$32 {v}^{3} + \left(64 - 12\right) {v}^{2} + \left(- 16 - 24\right) v - 32$

$32 {v}^{3} + 52 {v}^{2} + \left(- 40\right) v - 32$

$32 {v}^{3} + 52 {v}^{2} - 40 v - 32$