# How do you factor 10x^2 + 16x - 55x - 88?

Jun 17, 2016

$\left(5 x + 8\right) \left(2 x - 11\right)$

#### Explanation:

Group the four terms into twp groups.

$\left(10 {x}^{2} + 16 x\right) + \left(- 55 x - 88\right)$

$2 x \left(5 x + 8\right) - 11 \left(5 x + 8\right) \text{ take out common factors}$

$\left(5 x + 8\right) \left(2 x - 11\right) \text{ common bracket as a factor}$

Jun 17, 2016

$\left(5 x + 8\right) \left(2 x + 1\right)$

#### Explanation:

In the given polynomial we recognize having $2 x$ as common factor between $10 {x}^{2} \mathmr{and} 16 x$
Hence, $10 {x}^{2} + 16 x = 2 x \left(5 x + 8\right)$

$- 11$is the common factor of $- 55 x \mathmr{and} - 88$

$- 55 x - 88 = - 11 \left(5 x + 8\right)$

$10 {x}^{2} + 16 x - 55 x - 88$
$= 2 x \left(5 x + 8\right) - 11 \left(5 x + 8\right)$
Taking the common factor $5 x + 8$

$= \left(5 x + 8\right) \left(2 x - 11\right)$