Question #62fe0

Jan 20, 2018

$2 {x}^{2} - 18 = 2 \left(x + 3\right) \left(x - 3\right)$

Explanation:

First, we can bring out a $2$ since both terms are a multiple of $2$:
$2 {x}^{2} - 18 = 2 \left({x}^{2} - 9\right)$

Next we realize that ${x}^{2} - 9$ is a difference of squares. We know that the formula for difference of squares is:
$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

This means we can factor our expression like so:
$2 \left(x + 3\right) \left(x - 3\right)$

Jan 20, 2018

$2 \left(x + 3\right) \left(x - 3\right)$

Explanation:

First, we know that $18 = 9 \cdot 2$:
$2 {x}^{2} - 9 \cdot 2$

Therefore, factor out the common term $2$:
$\textcolor{red}{2} {x}^{2} - 9 \cdot \textcolor{red}{2}$ to be equals to:

$2 \left({x}^{2} - 9\right)$

Then, we rewrite $9$ as ${3}^{2}$:

$2 \left({x}^{2} - {3}^{2}\right)$.

We will remove $2$ for now.
We will then apply the DOS (Difference of squares) rule:
${x}^{2} - {y}^{2} = \left(x + y\right) \left(x - y\right)$

Therefore, $\left({x}^{2} - {3}^{2}\right) = \left(x + 3\right) \left(x - 3\right)$

Put back the $2$:
$2 \left(x + 3\right) \left(x - 3\right)$

Jan 20, 2018

$2 \left(x - 3\right) \left(x + 3\right)$

Explanation:

It seems like there's nothing to be done here about the ${x}^{2}$. Let's start by factoring out the $2$ to see if anything is "hidden".

$2 {x}^{2} - 18 = 2 \left({x}^{2} - 9\right)$

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

There's an identity for a difference of squares:

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

On such problems, if "factoring out numbers" doesn't seem to work, maybe you'll have to use one of those.