# How do you factor x^2 + 8x - 19?

May 12, 2016

${x}^{2} + 8 x - 19 = \left(x + 4 - \sqrt{35}\right) \left(x + 4 + \sqrt{35}\right)$

#### Explanation:

Complete the square and use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x + 4\right)$ and $b = \sqrt{35}$ as follows:

${x}^{2} + 8 x - 19$

$= {\left(x + 4\right)}^{2} - {4}^{2} - 19$

$= {\left(x + 4\right)}^{2} - 16 - 19$

$= {\left(x + 4\right)}^{2} - 35$

$= {\left(x + 4\right)}^{2} - {\left(\sqrt{35}\right)}^{2}$

$= \left(\left(x + 4\right) - \sqrt{35}\right) \left(\left(x + 4\right) + \sqrt{35}\right)$

$= \left(x + 4 - \sqrt{35}\right) \left(x + 4 + \sqrt{35}\right)$