If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Algebra 1

Course: Algebra 1 > Unit 13

Lesson 8: Factoring quadratics with perfect squares
Math
Algebra 1
Quadratics: Multiplying & factoring
Factoring quadratics with perfect squares
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

Factoring quadratics: Perfect squares

Google Classroom
Learn how to factor quadratics that have the "perfect square" form. For example, write x²+6x+9 as (x+3)².
Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication.
In this article, we'll learn how to factor perfect square trinomials using special patterns. This reverses the process of squaring a binomial, so you'll want to understand that completely before proceeding.

Intro: Factoring perfect square trinomials

To expand any binomial, we can apply one of the following patterns.
  • left parenthesis, start color #11accd, a, end color #11accd, plus, start color #1fab54, b, end color #1fab54, right parenthesis, squared, equals, start color #11accd, a, end color #11accd, squared, plus, 2, start color #11accd, a, end color #11accd, start color #1fab54, b, end color #1fab54, plus, start color #1fab54, b, end color #1fab54, squared
  • left parenthesis, start color #11accd, a, end color #11accd, minus, start color #1fab54, b, end color #1fab54, right parenthesis, squared, equals, start color #11accd, a, end color #11accd, squared, minus, 2, start color #11accd, a, end color #11accd, start color #1fab54, b, end color #1fab54, plus, start color #1fab54, b, end color #1fab54, squared
Note that in the patterns, a and b can be any algebraic expression. For example, suppose we want to expand left parenthesis, x, plus, 5, right parenthesis, squared. In this case, start color #11accd, a, end color #11accd, equals, start color #11accd, x, end color #11accd and start color #1fab54, b, end color #1fab54, equals, start color #1fab54, 5, end color #1fab54, and so we get:
(x+5)2=x2+2(x)(5)+(5)2=x2+10x+25\begin{aligned}(\blueD x+\greenD 5)^2&=\blueD x^2+2(\blueD x)(\greenD5)+(\greenD 5)^2\\\\ &=x^2+10x+25\end{aligned}
You can check this pattern by using multiplication to expand left parenthesis, x, plus, 5, right parenthesis, squared.
The reverse of this expansion process is a form of factoring. If we rewrite the equations in the reverse order, we will have patterns for factoring polynomials of the form a, squared, plus minus, 2, a, b, plus, b, squared.
  • start color #11accd, a, end color #11accd, squared, plus, 2, start color #11accd, a, end color #11accd, start color #1fab54, b, end color #1fab54, plus, start color #1fab54, b, end color #1fab54, squared, space, equals, left parenthesis, start color #11accd, a, end color #11accd, plus, start color #1fab54, b, end color #1fab54, right parenthesis, squared
  • start color #11accd, a, end color #11accd, squared, minus, 2, start color #11accd, a, end color #11accd, start color #1fab54, b, end color #1fab54, plus, start color #1fab54, b, end color #1fab54, squared, space, equals, left parenthesis, start color #11accd, a, end color #11accd, minus, start color #1fab54, b, end color #1fab54, right parenthesis, squared
We can apply the first pattern to factor x, squared, plus, 10, x, plus, 25. Here we have start color #11accd, a, end color #11accd, equals, start color #11accd, x, end color #11accd and start color #1fab54, b, end color #1fab54, equals, start color #1fab54, 5, end color #1fab54.
x2+10x+25=x2+2(x)(5)+(5)2=(x+5)2\begin{aligned}x^2+10x+25&=\blueD x^2+2(\blueD x)(\greenD5)+(\greenD 5)^2\\\\ &=(\blueD x+\greenD 5)^2\end{aligned}
Expressions of this form are called perfect square trinomials. The name reflects the fact that this type of three termed polynomial can be expressed as a perfect square!
Let's take a look at a few examples in which we factor perfect square trinomials using this pattern.

Example 1: Factoring x, squared, plus, 8, x, plus, 16

Notice that both the first and last terms are perfect squares: x, squared, equals, left parenthesis, start color #11accd, x, end color #11accd, right parenthesis, squared and 16, equals, left parenthesis, start color #1fab54, 4, end color #1fab54, right parenthesis, squared. Additionally, notice that the middle term is two times the product of the numbers that are squared: 2, left parenthesis, start color #11accd, x, end color #11accd, right parenthesis, left parenthesis, start color #1fab54, 4, end color #1fab54, right parenthesis, equals, 8, x.
This tells us that the polynomial is a perfect square trinomial, and so we can use the following factoring pattern.
start color #11accd, a, end color #11accd, squared, plus, 2, start color #11accd, a, end color #11accd, start color #1fab54, b, end color #1fab54, plus, start color #1fab54, b, end color #1fab54, squared, space, equals, left parenthesis, start color #11accd, a, end color #11accd, plus, start color #1fab54, b, end color #1fab54, right parenthesis, squared
In our case, start color #11accd, a, end color #11accd, equals, start color #11accd, x, end color #11accd and start color #1fab54, b, end color #1fab54, equals, start color #1fab54, 4, end color #1fab54. We can factor our polynomial as follows:
x2+8x+16=(x)2+2(x)(4)+(4)2=(x+4)2\begin{aligned}x^2+8x+16&=(\blueD x)^2+2(\blueD x)(\greenD 4)+(\greenD4)^2\\ \\ &=(\blueD{x}+\greenD{4})^2\end{aligned}
We can check our work by expanding left parenthesis, x, plus, 4, right parenthesis, squared:
(x+4)2=(x)2+2(x)(4)+(4)2=x2+8x+16\begin{aligned}(x+4)^2&=(x)^2+2(x)(4)+(4)^2\\ \\ &=x^2+8x+16 \end{aligned}

Check your understanding

1) Factor x, squared, plus, 6, x, plus, 9.
Choose 1 answer:

2) Factor x, squared, minus, 6, x, plus, 9.
Choose 1 answer:

3) Factor x, squared, plus, 14, x, plus, 49.

Example 2: Factoring 4, x, squared, plus, 12, x, plus, 9

It is not necessary for the leading coefficient of a perfect square trinomial to be 1.
For example, in 4, x, squared, plus, 12, x, plus, 9, notice that both the first and last terms are perfect squares: 4, x, squared, equals, left parenthesis, start color #11accd, 2, x, end color #11accd, right parenthesis, squared and 9, equals, left parenthesis, start color #1fab54, 3, end color #1fab54, right parenthesis, squared. Additionally, notice that the middle term is two times the product of the numbers that are squared: 2, left parenthesis, start color #11accd, 2, x, end color #11accd, right parenthesis, left parenthesis, start color #1fab54, 3, end color #1fab54, right parenthesis, equals, 12, x.
Because it satisfies the above conditions, 4, x, squared, plus, 12, x, plus, 9 is also a perfect square trinomial. We can again apply the following factoring pattern.
start color #11accd, a, end color #11accd, squared, plus, 2, start color #11accd, a, end color #11accd, start color #1fab54, b, end color #1fab54, plus, start color #1fab54, b, end color #1fab54, squared, space, equals, left parenthesis, start color #11accd, a, end color #11accd, plus, start color #1fab54, b, end color #1fab54, right parenthesis, squared
In this case, start color #11accd, a, end color #11accd, equals, start color #11accd, 2, x, end color #11accd and start color #1fab54, b, end color #1fab54, equals, start color #1fab54, 3, end color #1fab54. The polynomial factors as follows:
4x2+12x+9=(2x)2+2(2x)(3)+(3)2=(2x+3)2\begin{aligned}4x^2+12x+9&=(\blueD {2x})^2+2(\blueD {2x})(\greenD 3)+(\greenD3)^2\\ \\ &=(\blueD{2x}+\greenD{3})^2\end{aligned}
We can check our work by expanding left parenthesis, 2, x, plus, 3, right parenthesis, squared.

Check your understanding

4) Factor 9, x, squared, plus, 30, x, plus, 25.
Choose 1 answer:

5) Factor 4, x, squared, minus, 20, x, plus, 25.

Challenge problems

6*) Factor x, start superscript, 4, end superscript, plus, 2, x, squared, plus, 1.

7*) Factor 9, x, squared, plus, 24, x, y, plus, 16, y, squared.

Want to join the conversation?

Log in