# How do you multiply (5y+3)(4y^2-7y-2)?

Dec 14, 2016

$20 {y}^{3} - 23 {y}^{2} - 31 y - 6$

#### Explanation:

We must ensure that each term in the second bracket is multiplied by each term in the first bracket.

$\Rightarrow \left(\textcolor{red}{5 y + 3}\right) \left(4 {y}^{2} - 7 y - 2\right)$

$= \textcolor{red}{5 y} \left(4 {y}^{2} - 7 y - 2\right) \textcolor{red}{+ 3} \left(4 {y}^{2} - 7 y - 2\right)$

$\text{now distribute the brackets}$

$= 20 {y}^{3} - 35 {y}^{2} - 10 y + 12 {y}^{2} - 21 y - 6$

$\text{collecting like terms gives}$

$= 20 {y}^{3} + {y}^{2} \left(- 35 + 12\right) + y \left(- 10 - 21\right) - 6$

$= 20 {y}^{3} - 23 {y}^{2} - 31 y - 6$