Main content

### Course: Algebra 1 > Unit 13

Lesson 8: Factoring quadratics with perfect squares- Perfect square factorization intro
- Factoring quadratics: Perfect squares
- Perfect squares intro
- Factoring perfect squares
- Identifying perfect square form
- Factoring perfect squares: negative common factor
- Factoring perfect squares: missing values
- Factoring perfect squares: shared factors
- Perfect squares

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Factoring quadratics: 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.

$({a}+{b}{)}^{2}={{a}}^{2}+2{a}{b}+{{b}}^{2}$ $({a}-{b}{)}^{2}={{a}}^{2}-2{a}{b}+{{b}}^{2}$

Note that in the patterns, $a$ and $b$ can be any algebraic expression. For example, suppose we want to expand $(x+5{)}^{2}$ . In this case, ${a}={x}$ and ${b}={5}$ , and so we get:

You can check this pattern by using multiplication to expand $(x+5{)}^{2}$ .

The reverse of this expansion process is a form of ${a}^{2}\pm 2ab+{b}^{2}$ .

**factoring.**If we rewrite the equations in the reverse order, we will have patterns for factoring polynomials of the form${{a}}^{2}+2{a}{b}+{{b}}^{2}\text{}=({a}+{b}{)}^{2}$ ${{a}}^{2}-2{a}{b}+{{b}}^{2}\text{}=({a}-{b}{)}^{2}$

We can apply the first pattern to factor ${x}^{2}+10x+25$ . Here we have ${a}={x}$ and ${b}={5}$ .

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 ${x}^{2}+8x+16$

Notice that both the first and last terms are perfect squares: ${x}^{2}=({x}{)}^{2}$ and $16=({4}{)}^{2}$ . Additionally, notice that the middle term is two times the product of the numbers that are squared: $2({x})({4})=8x$ .

This tells us that the polynomial is a perfect square trinomial, and so we can use the following factoring pattern.

In our case, ${a}={x}$ and ${b}={4}$ . We can factor our polynomial as follows:

We can check our work by expanding $(x+4{)}^{2}$ :

### Check your understanding

## Example 2: Factoring $4{x}^{2}+12x+9$

It is not necessary for the leading coefficient of a perfect square trinomial to be $1$ .

For example, in $4{x}^{2}+12x+9$ , notice that both the first and last terms are perfect squares: $4{x}^{2}=({2x}{)}^{2}$ and $9=({3}{)}^{2}$ . Additionally, notice that the middle term is two times the product of the numbers that are squared: $2({2x})({3})=12x$ .

Because it satisfies the above conditions, $4{x}^{2}+12x+9$ is also a perfect square trinomial. We can again apply the following factoring pattern.

In this case, ${a}={2x}$ and ${b}={3}$ . The polynomial factors as follows:

We can check our work by expanding $(2x+3{)}^{2}$ .

### Check your understanding

## Challenge problems

## Want to join the conversation?

- is (4+x)^2 the same as (x+4)^2(26 votes)
- Yes, they are the same, as you can switch the numbers around within the parenthesis, just not outside of them.(39 votes)

- Just curious, what's the point of doing this to an expression? Like, where will this be applied?(11 votes)
- 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!(42 votes)

- Number three marks (X+7) squared wrong as well, but the "I need help" section is correct.(4 votes)
- Maybe your Caps Lock is on. (X+7) is different from (x+7).

Hope I have helped :)(26 votes)

- 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(7 votes)
- 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.(11 votes)

- What do the a and b stand for in the equation a^2+2ab+b^2 ? I'm very confused :/(4 votes)
- they are just any varibles. a and b can be anything (a has to be equal to a and b has to be equal to b tho)(1 vote)

- is algebra 1 a high school class because im taking it in middle school(3 votes)
- 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(5 votes)

- Does anyone understand this at all?(5 votes)
- It was overwhelming for me at first, and it’s hard to keep track of all the different methods, but something that helps me is going back and walking through the steps when I get confused. Take one small chunk at a time and make sure it makes sense, then move on to the next one. You got this! It’s ok if it’s slow going.(1 vote)

- problem 7 doesnt have power (degree) button to write the correct answer. Need Fix(3 votes)
- No fix is needed. You can use the ^ symbols to enter an exponent. (3x+4y)^2

Alternatively, you can enter it as (3x+4y)(3x+4y)(5 votes)

- so like do we do the stuff we learned before or ?? what are we doing??(4 votes)
- For question 7, [9x^2+24xy+16y^2]. The first and last terms are perfect squares (3x and 4y). But when I express the middle term like this, 2(3x)(4y) I got 6x8y instead of 24xy. Can someone explain?(2 votes)
- You have a multiplication error. There is no distributive property that distributes multiplication across multiplication. The distributive property only distributes multiplication across addition or subtraction.

To multiply 3 or more items, you do them in pairs.

2(3x) = 6x

6x(4y) = 24xy

Hope this helps.(4 votes)