# How do you find the product (4y^2-3)(4y^2+7y+2)?

Mar 12, 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.

$\left(\textcolor{red}{4 {y}^{2}} - \textcolor{red}{3}\right) \left(\textcolor{b l u e}{4 {y}^{2}} + \textcolor{b l u e}{7 y} + \textcolor{b l u e}{2}\right)$ becomes:

$\left(\textcolor{red}{4 {y}^{2}} \times \textcolor{b l u e}{4 {y}^{2}}\right) + \left(\textcolor{red}{4 {y}^{2}} \times \textcolor{b l u e}{7 y}\right) + \left(\textcolor{red}{4 {y}^{2}} \times \textcolor{b l u e}{2}\right) - \left(\textcolor{red}{3} \times \textcolor{b l u e}{4 {y}^{2}}\right) - \left(\textcolor{red}{3} \times \textcolor{b l u e}{7 y}\right) - \left(\textcolor{red}{3} \times \textcolor{b l u e}{2}\right)$

$16 {y}^{4} + 28 {y}^{3} + 8 {y}^{2} - 12 {y}^{2} - 21 y - 6$

We can now combine like terms:

$16 {y}^{4} + 28 {y}^{3} + \left(8 - 12\right) {y}^{2} - 21 y - 6$

$16 {y}^{4} + 28 {y}^{3} - 4 {y}^{2} - 21 y - 6$