# How do you multiply (2x^2+6x-8)(2x^2-6x-3)?

Jun 30, 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.

$\left(\textcolor{red}{2 {x}^{2}} + \textcolor{red}{6 x} - \textcolor{red}{8}\right) \left(\textcolor{b l u e}{2 {x}^{2}} - \textcolor{b l u e}{6 x} - \textcolor{b l u e}{3}\right)$ becomes:

$\left(\textcolor{red}{2 {x}^{2}} \times \textcolor{b l u e}{2 {x}^{2}}\right) - \left(\textcolor{red}{2 {x}^{2}} \times \textcolor{b l u e}{6 x}\right) - \left(\textcolor{red}{2 {x}^{2}} \times \textcolor{b l u e}{3}\right) + \left(\textcolor{red}{6 x} \times \textcolor{b l u e}{2 {x}^{2}}\right) - \left(\textcolor{red}{6 x} \times \textcolor{b l u e}{6 x}\right) - \left(\textcolor{red}{6 x} \times \textcolor{b l u e}{3}\right) - \left(\textcolor{red}{8} \times \textcolor{b l u e}{2 {x}^{2}}\right) - \left(\textcolor{red}{8} \times \textcolor{b l u e}{6 x}\right) + \left(\textcolor{red}{8} \times \textcolor{b l u e}{3}\right)$

$4 {x}^{4} - 12 {x}^{3} - 6 {x}^{2} + 12 {x}^{3} - 36 {x}^{2} - 18 x - 16 {x}^{2} - 48 x + 24$

We can now group and combine like terms:

$4 {x}^{4} - 12 {x}^{3} + 12 {x}^{3} - 6 {x}^{2} - 36 {x}^{2} - 16 {x}^{2} - 18 x - 48 x + 24$

$4 {x}^{4} + \left(- 12 + 12\right) {x}^{3} + \left(- 62 - 36 - 16\right) {x}^{2} + \left(- 18 - 48\right) x + 24$

$4 {x}^{4} + \left(0\right) {x}^{3} + \left(- 114\right) {x}^{2} + \left(- 66\right) x + 24$

$4 {x}^{4} - 114 {x}^{2} - 66 x + 24$