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## Algebra 1 (Eureka Math/EngageNY)

### Course: Algebra 1 (Eureka Math/EngageNY) > Unit 4

Lesson 1: Topic A: Lessons 1-2: Factoring monomials- Intro to factors & divisibility
- Intro to factors & divisibility
- Factors & divisibility
- Which monomial factorization is correct?
- Factoring monomials
- Worked example: finding the missing monomial factor
- Worked example: finding missing monomial side in area model
- Factor monomials

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# Factoring monomials

Learn how to completely factor monomial expressions, or find the missing factor in a monomial factorization.

#### What you should be familiar with before this lesson

A $x$ , like $3{x}^{2}$ . A $3{x}^{2}+6x-1$ .

**monomial**is an expression that is the product of constants and nonnegative integer powers of**polynomial**is a sum of monomials, likeIf $A=B\cdot C$ , then $B$ and $C$ are $A$ , and $A$ is $B$ and $C$ . To review this material, check out our article on Factoring and divisibility.

**factors**of**divisible**by#### What you will learn in this lesson

In this lesson, you will learn how to factor monomials. You will use what you already know about factoring integers to help you in this quest.

## Introduction: What is monomial factorization?

To

**factor**a monomial means to express it as a product of two or more monomials.For example, below are several possible factorizations of $8{x}^{5}$ .

$8{x}^{5}=(2{x}^{2})(4{x}^{3})$ $8{x}^{5}=(8x)({x}^{4})$ $8{x}^{5}=(2x)(2x)(2x)({x}^{2})$

Notice that when you multiply each expression on the right, you get $8{x}^{5}$ .

### Reflection question

## Completely factoring monomials

#### Review: integer factorization

To factor an integer completely, we write it as a product of primes.

For example, we know that $30=2\cdot 3\cdot 5$ .

#### And now to monomials...

To factor a monomial completely, we write the coefficient as a product of primes

*and*expand the variable part.For example, to completely factor $10{x}^{3}$ , we can write the prime factorization of $10$ as $2\cdot 5$ and write ${x}^{3}$ as $x\cdot x\cdot x$ . Therefore, this is the complete factorization of $10{x}^{3}$ :

### Check your understanding

## Finding missing factors of monomials

#### Review: integer factorization

Suppose we know that $56=8b$ for some integer $b$ . How can we find the other factor?

Well, we can solve the equation $56=8b$ for $b$ by dividing both sides of the equation by $8$ . The missing factor is $7$ .

#### And now to monomials...

We can extend these ideas to monomials. For example, suppose $8{x}^{5}=(4{x}^{3})(C)$ for some monomial $C$ . We can find $C$ by dividing $8{x}^{5}$ by $4{x}^{3}$ :

We can check our work by showing that the product of $4{x}^{3}$ and $2{x}^{2}$ is indeed $8{x}^{5}$ .

### Check your understanding

## A note about multiple factorizations

Consider the number $12$ . We can write four different factorizations of this number.

$12=2\cdot 6$ $12=3\cdot 4$ $12=12\cdot 1$ $12=2\cdot 2\cdot 3$

However, there is only $12$ , i.e. $2\cdot 2\cdot 3$ .

*one*prime factorization of the numberThe same idea holds with monomials. We can factor $18{x}^{3}$ in many ways. Here are a few different factorizations.

$18{x}^{3}=2\cdot 9\cdot {x}^{3}$ $18{x}^{3}=3\cdot 6\cdot x\cdot {x}^{2}$ $18{x}^{3}=2\cdot 3\cdot 3\cdot {x}^{3}$

Yet there is only one complete factorization!

## Challenge problems

## Want to join the conversation?

- i need help with this i have already tried a lot but i just don't get it.(16 votes)
- Let me see if I can help.

One way to look at factorizing monomials is pulling them apart into their basic parts. Like you had a peanut butter and jelly sandwich. Pulling that into its basic parts, you could get bread times bread times peanut butter times jelly. That would be its complete factorization.

Now for a monomial

Say you have a monomial (or one term) that is 24x^2. There are a few ways to factor this. One example would be: (4x)(6x) This is not a complete factorization, but it is still a valid way to factorize it. Another way to do it would be (3)(8x^2) Its complete factorization would be 2 times 2 times 2 times 3 times x times x.

When factoring, you just need to make sure that when you preform all the operations, they still come out to be the original number. Like I need to make sure (4x)(6x) = 24x^2. Let's check. 4 times 6 equals 24, and x times x equals x^2. Those two multiplied together equals 24x^2.

I hope that helped! :)(32 votes)

- You don't earn energy points on the articles, but you guys do it anyway. Good job! Keep it up.(14 votes)
- yeah, i mean i noticed its just summary material-not new stuff(7 votes)

- I'm having a hard time understanding how to find the width of the rectangle,how should i go about finding the width of a rectangle?(6 votes)
- Length times width equals area. So to find width, you would divide the area by length.

hope that helps!(15 votes)

- I'm so glad that he went into more detail with more videos. Because in the first video I was so lost lol.(12 votes)
- I am still confused on how to tell if it is a monomial or not. Can you help me?(8 votes)
- A monomial is a single term polynomial, with a non-negative integer exponent.

Typically each term in a polynomial is separated by addition/subtraction.(8 votes)

- If it has 1 term, it is a monomial. If it has 2 terms, it is a binomial. 3 terms or more is a polynomial.(4 votes)
- Slight clarification: All of them are polynomials. The classification as monomial, binomial, or trinomial just gives more descriptive information about the polynomial. You omitted trinomial from you list. It has 3 terms.(13 votes)

- How can understand this better?(1 vote)
- as i tell myself practice makes progress not perfect because you cant be perfect or you'll die trying!(11 votes)

- this helped a lot but is there anyway to understand it better(4 votes)
- search Youtube for extra explanations(2 votes)

- idk what most of these are(4 votes)
- Well, one way you can complete factorization is when you list the prime factors (they can't be simplified anymore) along with the amount of x's. For example, the complete factorization of 64x^6 would work like this:

64 = 8*8 = 2*2*2*2*2*2 (6 2's)

so you would write the complete factorization as: 2*2*2*2*2*2*x*x*x*x*x*x, this is 6 2's and 6 x's.

A complete factorization is one where the factors cannot be simplified anymore, like shown above. In addition, there is only one complete factorization for a given monomial.

If you don't have to completely factor the monomial, you can first find factors of the coefficient and then split the x's. For example, if I had 64x^6 again, I could list out the factors of 64 (coefficient) first.

64: 1 & 64, 2 & 32, 4 & 16, 8 & 8

You could then pick any of these pairs. Let's say we use 4 and 16. Then, we factor x^6, we could turn this into any two pairs that add up to 6. This is because of the exponent multiplication method. When you have the same base (x), you can add the exponents.

6 = 1+5, 2+4, 3+3. Let's say we choose 3+3.

Now, we can write it out. Recall that our coefficients are 4 and 16.

We can write it out like this: 4x^3 * 16x^3.

If we check it, we can multiply the coefficients (16*4 = 64) and the x's (x^3 * x^3 = x^6) and we get 64x^6, so this factorization is correct. Note that this is not the only factorization of 64x^6, you could select any other group of x's and any other coefficients that work.

I hope this helps and let me know if you have any questions! :)

P.S. sorry for the late reply(2 votes)

- philosophically why do we simplify?(2 votes)
- if someone told you, "I want 2 bottles of water, then 5 bottles of water, then one bottle of water, then 1 bottle of water!" and you say, "whats that total?" your asking them to simplify because its easier.

sim·pli·fy

/ˈsimpləˌfī/

verb

make (something) simpler or easier to do or understand.

"an overhaul of court procedure to simplify litigation"(1 vote)