# How do you find the product (4a+7)(9a^2+2a-7)?

Apr 18, 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.

$\left(\textcolor{red}{4 a} + \textcolor{red}{7}\right) \left(\textcolor{b l u e}{9 {a}^{2}} + \textcolor{b l u e}{2 a} - \textcolor{b l u e}{7}\right)$ becomes:

$\left(\textcolor{red}{4 a} \times \textcolor{b l u e}{9 {a}^{2}}\right) + \left(\textcolor{red}{4 a} \times \textcolor{b l u e}{2 a}\right) - \left(\textcolor{red}{4 a} \times \textcolor{b l u e}{7}\right) + \left(\textcolor{red}{7} \times \textcolor{b l u e}{9 {a}^{2}}\right) + \left(\textcolor{red}{7} \times \textcolor{b l u e}{2 a}\right) - \left(\textcolor{red}{7} \times \textcolor{b l u e}{7}\right)$

$36 {a}^{3} + 8 {a}^{2} - 28 a + 63 {a}^{2} + 14 a - 49$

We can now group and combine like terms:

$36 {a}^{3} + 8 {a}^{2} + 63 {a}^{2} - 28 a + 14 a - 49$

$36 {a}^{3} + \left(8 + 63\right) {a}^{2} + \left(- 28 + 14\right) a - 49$

$36 {a}^{3} + 71 {a}^{2} + \left(- 14\right) a - 49$

$36 {a}^{3} + 71 {a}^{2} - 14 a - 49$