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## Algebra 1

### Course: Algebra 1>Unit 13

Lesson 8: Factoring quadratics with perfect squares

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:
\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.
\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 $x^2+8x+16$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:
\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:
\begin{aligned}(x+4)^2&=(x)^2+2(x)(4)+(4)^2\\ \\ &=x^2+8x+16 \end{aligned}

1) Factor x, squared, plus, 6, x, plus, 9.

2) Factor x, squared, minus, 6, x, plus, 9.

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

## Example 2: Factoring $4x^2+12x+9$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:
\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.

4) Factor 9, x, squared, plus, 30, x, plus, 25.

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?

• is (4+x)^2 the same as (x+4)^2
• Yes, they are the same, as you can switch the numbers around within the parenthesis, just not outside of them.
• Just curious, what's the point of doing this to an expression? Like, where will this be applied?
• Pretty much all of algebra will be used when we start doing calculus which is where math gets really cool with way more real world applications like launching satellites into space and making vaccines. Endless possibilities! But we must have a SOLID foundation in algebra for all this cool stuff!
• Number three marks (X+7) squared wrong as well, but the "I need help" section is correct.
• Maybe your Caps Lock is on. (X+7) is different of (x+7).
Hope I have helped :)
• I don't understand question 4 and 5. in question 4 he said ""notice that the middle term is two times the product of the numbers that are squared: 2(3x)(5)=30x2(3x})(\D 5)=30x2(3x)(5)=30x2, l, 3, x, D, 5, equals, 30, x. then he does the equation again and shows the answer I don't understand how he got there
• normally when you have for example x^2+6x+9 you would take the root of 9 and multiply it by 2. if the answer is the first-degree variable coefficient (6) then it satisfy the perfect squares "trick" rule, and the answer would be (x+3)^2.

when you have a coefficient to the 2nd-degree variable like in 9x^2+30x+25, you do almost the same:
first you find the root of the 0-degree variable (25), which is 5. (5^2=25)
then you find the root of the 2nd-degree variable's coefficient (9), which is 3 (3^2=9).

now here is the different from before: you need to multiply both roots (3 and 5) and then multiply them by 2 (as you normally would). if the answer is the same as the 1st-degree variable's coefficient ((5*3)*2=30) then the trick can be applied here also, giving you the answer made of those 2 roots you found erlier(3 and 5)-squared.
(3x+5)^2
hope you understood and that it helped you.
• What do the a and b stand for in the equation a^2+2ab+b^2 ? I'm very confused :/
• "A" represents the variable or "x" in the problem. The "b" represents the y. They are the same and just mean that you can stick a number in their places.
• is algebra 1 a high school class because im taking it in middle school
• It can be, many people take algebra in middle school but there are also plenty of people who take it in high school. I think most people take geometry freshman year in high school (9th grade) but there are plenty of people who take it before or after
• I just factored 16x^2+80x+100 using the perfect square trinomial method and got (4x+10)^2. However, the answer key to the worksheet said that the answer is 4(2x+5)^2. Both answers are equal to 16x^2+80x+100. What method was used to get 4(2x+5)^2, and how do I know what method to use over the other?
(1 vote)
• Your answer is not correct because it is not in it's simplest form. First, you must find the greatest common factor, which would be 4 in this situation. So, 16x^2+80x+100= 4(4x^2+20x+25). Then, 4(4x^2+20x+25) can be further factored to equal 4(2x+5)^2. I hope this helps.
• in problem 5 we can write (-2x+5)^2. i wonder why this works.