# How do you factor x^{2}-8x=2?

Oct 13, 2017

$x = 4 \pm 3 \sqrt{2}$

#### Explanation:

I'm assuming we need to also solve the equation.

The first step is to move all the terms to one side.

${x}^{2} - 8 x = 2$

${x}^{2} - 8 x - 2 = 0$

Now, to factor this, we need to find two factors of $- 2$ which add up to $- 8$

Through a little bit of trial and error, you will find that there actually aren't any pairs of factors of $- 2$ that add up to $- 8$. We're going to have to solve this another way.

Let's try completing the square. We know that a perfect square has the form:

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

In this case, we have the following:

$2 a x = - 8 x$
$2 a = - 8$
$a = - 4$
${a}^{2} = 16$

To get $\textcolor{b l u e}{16}$, we need to add and subtract it from the left side, so it cancels out, like this:

${x}^{2} - 8 x - 2 = 0$

${x}^{2} - 8 x + \textcolor{b l u e}{16} - \textcolor{b l u e}{16} - 2 = 0$

$\left({x}^{2} - 8 x + 16\right) - 18 = 0$

${\left(x - 4\right)}^{2} - 18 = 0$

Now we can continue to solve the equation:

${\left(x - 4\right)}^{2} = 18$

$\left(x - 4\right) = \pm \sqrt{18}$

$x = 4 \pm \sqrt{18}$

This is our solution! Just one final adjustment to make. Remember that $18 = 2 \cdot 9 = 2 \cdot {3}^{2}$.

$x = 4 \pm \sqrt{2 \cdot {3}^{2}}$

$x = 4 \pm 3 \sqrt{2}$