# How do you factor 3y²-y+9y-3?

Find the common factor in the first 2 terms that is the same as the second 2 terms, and arrive at $\left(3 y - 1\right) \left(y + 3\right)$

#### Explanation:

This problem is an a form where the first 2 terms and the second 2 terms have a common factor, so let's take advantage of that.

$3 {y}^{2} - y + 9 y - 3$

Let's factor out a $y$ from the first 2 terms and a 3 from the second two terms (you'll learn to recognize these patterns with practice):

$y \left(3 y - 1\right) + 3 \left(3 y - 1\right)$

We can take the common factor and factor it out, like this:

$\left(3 y - 1\right) \left(y + 3\right)$