# How do you simplify (x - 1)(2x + 1)(x + 5)?

Jun 20, 2017

See a solution process below:

#### Explanation:

To simplify this expression, first, multiply the two terms on the left of the expression. To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

$\left(x - 1\right) \left(\textcolor{red}{2 x} + \textcolor{red}{1}\right) \left(\textcolor{b l u e}{x} + \textcolor{b l u e}{5}\right)$ becomes:

$\left(x - 1\right) \left(\left(\textcolor{red}{2 x} \times \textcolor{b l u e}{x}\right) + \left(\textcolor{red}{2 x} \times \textcolor{b l u e}{5}\right) + \left(\textcolor{red}{1} \times \textcolor{b l u e}{x}\right) + \left(\textcolor{red}{1} \times \textcolor{b l u e}{5}\right)\right)$

$\left(x - 1\right) \left(2 {x}^{2} + 10 x + x + 5\right)$

We can now combine like terms:

$\left(x - 1\right) \left(2 {x}^{2} + 10 x + 1 x + 5\right)$

$\left(x - 1\right) \left(2 {x}^{2} + \left(10 + 1\right) x + 5\right)$

$\left(x - 1\right) \left(2 {x}^{2} + 11 x + 5\right)$

To complete the simplification complete the same process on the two remaining terms:

$\left(\textcolor{red}{x} - \textcolor{red}{1}\right) \left(\textcolor{b l u e}{2 {x}^{2}} + \textcolor{b l u e}{11 x} + \textcolor{b l u e}{5}\right)$ becomes:

$\left(\textcolor{red}{x} \times \textcolor{b l u e}{2 {x}^{2}}\right) + \left(\textcolor{red}{x} \times \textcolor{b l u e}{11 x}\right) + \left(\textcolor{red}{x} \times \textcolor{b l u e}{5}\right) - \left(\textcolor{red}{1} \times \textcolor{b l u e}{2 {x}^{2}}\right) - \left(\textcolor{red}{1} \times \textcolor{b l u e}{11 x}\right) - \left(\textcolor{red}{1} \times \textcolor{b l u e}{5}\right)$

$2 {x}^{3} + 11 {x}^{2} + 5 x - 2 {x}^{2} - 11 x - 5$

We can now group and combine like terms:

$2 {x}^{3} + 11 {x}^{2} - 2 {x}^{2} + 5 x - 11 x - 5$

$2 {x}^{3} + \left(11 - 2\right) {x}^{2} + \left(5 - 11\right) x - 5$

$2 {x}^{3} + 9 {x}^{2} + \left(- 6\right) x - 5$

$2 {x}^{3} + 9 {x}^{2} - 6 x - 5$