# How do you solve 3x^2-27=0 by factoring?

Mar 19, 2018

$x = \pm 3$

#### Explanation:

We can factor out a $3$ from both terms. Here, we are essentially dividing. We get:

$3 \textcolor{b l u e}{\left({x}^{2} - 9\right)} = 0$

What I have in blue is called a Difference of Squares. What this means is that if I have the binomial

${a}^{2} - {b}^{2}$

Then it can be factored as $\left(a + b\right) \left(a - b\right)$

In our example, $a$ would be $x$ (square root of ${x}^{2}$) and $b$ would be $3$ (square root of $9$).

Since $a = x$ and $b = 3$, we have

$3 {\underbrace{\left(x + 3\right) \left(x - 3\right)}}_{{x}^{2} - 9} = 0$

Setting each factor equal to zero, we get:

$x = 3$ and $x = - 3$, or alternatively, $x = \pm 3$

If the concept of a "Difference of Squares" seems foreign to you, I encourage you to Google it or search it on Khan Academy to make sure you understand it.

Hope this helps!