# How do you simplify (2r -5)(r + 10)?

Sep 22, 2015

$2 {r}^{2} + 15 r - 50$

#### Explanation:

To do this we use a technique called FOIL. It stands for firsts, outers, inners, and lasts.

$\left(2 r - 5\right) \left(r + 10\right)$
Here, $2 r$ and $r$ are the firsts, $2 r$ and $10$ are outers, $- 5$ and $r$ are inners, and $- 5$ and $10$ are lasts. Solving the problem only requires that we multiply these pairs together and combine them.

$\left(2 r - 5\right) \left(r + 10\right)$
$\left(2 r \cdot r\right) + \left(2 r \cdot 10\right) + \left(- 5 \cdot r\right) + \left(- 5 \cdot 10\right)$
$2 {r}^{2} + 20 r - 5 r - 50$
$2 {r}^{2} + 15 r - 50$