# How do you factor x² - 4x - 45?

Jul 6, 2015

Find two factors of $45$ whose difference is $4$. The numbers $9$ and $5$ work.

Hence ${x}^{2} - 4 x - 45 = \left(x - 9\right) \left(x + 5\right)$

#### Explanation:

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

So given ${x}^{2} - 4 x - 45$, look for $a$ and $b$

such that $a b = 45$ and $a - b = 4$.

Jul 6, 2015

Alternatively, complete the square to find:

${x}^{2} - 4 x - 45 = {\left(x - 2\right)}^{2} - {7}^{2} =$

$\left(x - 2 - 7\right) \left(x - 2 + 7\right) = \left(x - 9\right) \left(x + 5\right)$

#### Explanation:

${x}^{2} - 4 x - 45$

$= {x}^{2} - 4 x + 4 - 4 - 45$

$= {\left(x - 2\right)}^{2} - 49$

$= {\left(x - 2\right)}^{2} - {7}^{2}$

This is a difference of squares, so we can use the identity:

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

Let $a = x - 2$ and $b = 7$

Then

${\left(x - 2\right)}^{2} - {7}^{2}$

$= \left(x - 2 - 7\right) \left(x - 2 + 7\right) = \left(x - 9\right) \left(x + 5\right)$