# How do you factor x^ { 2} + 6 x - 27= 0?

Sep 22, 2017

$\left(x - 3\right) \left(x + 9\right)$

#### Explanation:

To factor ${x}^{2} + 6 x - 27$

Find to numbers whose product is -27 and whose sum is 6. For this it is -3 and 9.

$- 3 \times 9 = - 27$ and $- 3 + 9 = 6$.

Rewrite the equation as:

${x}^{2} - 3 x + 9 x - 27$

Factor the first 2 terms:

$x \left(x - 3\right)$

Factor the last 2 terms:

$9 \left(x - 3\right)$

So now we have:

$x \left(x - 3\right) + 9 \left(x - 3\right)$

Notice we can bracket off this and factor out the expressions in parenthesis:

$\left[x \left(x - 3\right) + 9 \left(x - 3\right)\right]$

$\left(x - 3\right) \left[x + 9\right]$

These are our required factors:

$\left(x - 3\right) \left(x + 9\right)$

Sep 22, 2017

$\left(x + 9\right) \left(x - 3\right) = 0$

#### Explanation:

Given:

${x}^{2} + 6 x - 27$

Here are a couple of methods:

Method 1 - Fishing for factors

Find a pair of factors of $27$ with difference $6$.

The pair $9 , 3$ works in that $9 \cdot 3 = 27$ and $9 - 3 = 6$

So we find:

${x}^{2} + 6 x - 27 = \left(x + 9\right) \left(x - 3\right)$

Method 2 - Completing the square

${x}^{2} + 6 x - 27 = {x}^{2} + 6 x + 9 - 36$

$\textcolor{w h i t e}{{x}^{2} + 6 x - 27} = {\left(x + 3\right)}^{2} - {6}^{2}$

$\textcolor{w h i t e}{{x}^{2} + 6 x - 27} = \left(\left(x + 3\right) - 6\right) \left(\left(x + 3\right) + 6\right)$

$\textcolor{w h i t e}{{x}^{2} + 6 x - 27} = \left(x - 3\right) \left(x + 9\right)$