# How do you find the product of (3d+3)(2d^2+5d-2)?

Mar 25, 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.

$\left(\textcolor{red}{3 d} + \textcolor{red}{3}\right) \left(\textcolor{b l u e}{2 {d}^{2}} + \textcolor{b l u e}{5 d} - \textcolor{b l u e}{2}\right)$ becomes:

$\left(\textcolor{red}{3 d} \times \textcolor{b l u e}{2 {d}^{2}}\right) + \left(\textcolor{red}{3 d} \times \textcolor{b l u e}{5 d}\right) - \left(\textcolor{red}{3 d} \times \textcolor{b l u e}{2}\right) + \left(\textcolor{red}{3} \times \textcolor{b l u e}{2 {d}^{2}}\right) + \left(\textcolor{red}{3} \times \textcolor{b l u e}{5 d}\right) - \left(\textcolor{red}{3} \times \textcolor{b l u e}{2}\right)$

$6 {d}^{3} + 15 {d}^{2} - 6 d + 6 {d}^{2} + 15 d - 6$

We can now group and combine like terms:

$6 {d}^{3} + 15 {d}^{2} + 6 {d}^{2} - 6 d + 15 d - 6$

$6 {d}^{3} + \left(15 + 6\right) {d}^{2} + \left(- 6 + 15\right) d - 6$

$6 {d}^{3} + 21 {d}^{2} + 9 d - 6$