# Question #c7477

Feb 13, 2018

$2 {x}^{4} + 16 {x}^{3} + 35 {x}^{2} + 10 x - 21$

#### Explanation:

In problems like these, we have to multiply everything in the first set of parentheses by everything in the second parenthesis.

$\textcolor{b l u e}{{x}^{2}} + \textcolor{red}{6 x} + \textcolor{\mathmr{and} a n \ge}{7}$

$\textcolor{b l u e}{{x}^{2}} \left(2 {x}^{2} + 4 x - 3\right) + \textcolor{red}{6 x} \left(2 {x}^{2} + 4 x - 3\right) + \textcolor{\mathmr{and} a n \ge}{7} \left(2 {x}^{2} + 4 x - 3\right)$

Distributing the $\textcolor{b l u e}{{x}^{2}}$:
$2 {x}^{4} + 4 {x}^{3} - 3 {x}^{2}$

Distributing the $\textcolor{red}{6 x}$:
$12 {x}^{3} + 24 {x}^{2} - 18 x$

Distributing the $\textcolor{\mathmr{and} a n \ge}{7}$:
$14 {x}^{2} + 28 x - 21$

Putting it all together:
$2 {x}^{4} + 4 {x}^{3} - 3 {x}^{2} + 12 {x}^{3} + 24 {x}^{2} - 18 x + 14 {x}^{2} + 28 x - 21$

Combining like terms:

$2 {x}^{4} + 16 {x}^{3} + 35 {x}^{2} + 10 x - 21$