# How do you find the product (2y-11)(y^2-3y+2)?

May 9, 2017

See a 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.

$\left(\textcolor{red}{2 y} - \textcolor{red}{11}\right) \left(\textcolor{b l u e}{{y}^{2}} - \textcolor{b l u e}{3 y} + \textcolor{b l u e}{2}\right)$ becomes:

$\left(\textcolor{red}{2 y} \times \textcolor{b l u e}{{y}^{2}}\right) - \left(\textcolor{red}{2 y} \times \textcolor{b l u e}{3 y}\right) + \left(\textcolor{red}{2 y} \times \textcolor{b l u e}{2}\right) - \left(\textcolor{red}{11} \times \textcolor{b l u e}{{y}^{2}}\right) + \left(\textcolor{red}{11} \times \textcolor{b l u e}{3 y}\right) - \left(\textcolor{red}{11} \times \textcolor{b l u e}{2}\right)$

$2 {y}^{3} - 6 {y}^{2} + 4 y - 11 {y}^{2} + 33 y - 22$

We can now group and combine like terms:

$2 {y}^{3} - 6 {y}^{2} - 11 {y}^{2} + 4 y + 33 y - 22$

$2 {y}^{3} + \left(- 6 - 11\right) {y}^{2} + \left(4 + 33\right) y - 22$

$2 {y}^{3} + \left(- 17\right) {y}^{2} + 37 y - 22$

$2 {y}^{3} - 17 {y}^{2} + 37 y - 22$