Factoring quadratics is very similar to multiplying binomials, just going the other way. For example, x^2+3x+2 factors to (x+1)(x+2) because (x+1)(x+2) multiplies to x^2+3x+2. This article reviews the basics of how to factor quadratics into the product of two binomials.

### Example

Factor as the product of two binomials.
x, start superscript, 2, end superscript, plus, 3, x, plus, 2
Our goal is to rewrite the expression in the form:
left parenthesis, x, plus, a, right parenthesis, left parenthesis, x, plus, b, right parenthesis
Expanding left parenthesis, x, plus, a, right parenthesis, left parenthesis, x, plus, b, right parenthesis gives us a clue.
\begin{aligned} x^2+\goldD{3}x+\blueD{2}&=(x+a)(x+b) \\\\ &= x^2 +ax+bx + ab \\\\ &= x^2 +\goldD{(a+b)}x + \blueD{ab} \end{aligned}
So start color goldD, left parenthesis, a, plus, b, right parenthesis, equals, 3, end color goldD and start color blueD, a, b, equals, 2, end color blueD.
After playing around with different possibilities for a and b, we discover that a, equals, start color greenD, 1, end color greenD, b, equals, start color greenD, 2, end color greenD satisfies both conditions.
Plugging these in, we get:
left parenthesis, x, plus, start color greenD, 1, end color greenD, right parenthesis, left parenthesis, x, plus, start color greenD, 2, end color greenD, right parenthesis
And we can multiply the binomials to check our solution if we'd like:
\begin{aligned} &~(x+1)(x+2) \\\\ =&~x^2+2x+x+2 \\\\ =&~x^2+3x+2 \end{aligned}
Yep, we get our original expression back, so we know we factored correctly to get our answer:
left parenthesis, x, plus, start color greenD, 1, end color greenD, right parenthesis, left parenthesis, x, plus, start color greenD, 2, end color greenD, right parenthesis
Want to see another example? Check out this video.

## Practice

Factor the quadratic expression as the product of two binomials.
x, start superscript, 2, end superscript, minus, x, minus, 42, equals

Want more practice? Check out this exercise.