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### 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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# Intro to factors & divisibility

Learn what it means for polynomials to be factors of other polynomials or to be divisible by them.

## What we need to know for 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 an expression that consists of a sum of monomials, like## What we will learn in this lesson

In this lesson, we will explore the relationship between factors and divisibility in polynomials and also learn how to determine if one polynomial is a factor of another.

## Factors and divisibility in integers

In general, two integers that multiply to obtain a number are considered

**factors**of that number.For example, since $14=2\cdot 7$ , we know that $2$ and $7$ are $14$ .

**factors**ofOne number is

**divisible**by another number if the result of the division is an integer.For example, since $\frac{15}{3}}=5$ and $\frac{15}{5}}=3$ , then $15$ is divisible by $3$ and $5$ . However, since $\frac{9}{4}}=2.25$ , then $9$ is $4$ .

*not divisible*byNotice the mutual relationship between factors and divisibility:

Since ${14}={2}\cdot 7$ (which means $2$ is a factor of $14$ ), we know that $\frac{{14}}{{2}}}=7$ (which means $14$ is divisible by $2$ ).

In the other direction, since $\frac{{15}}{{3}}}=5$ (which means $15$ is divisible by $3$ ), we know that ${15}={3}\cdot 5$ (which means $3$ is a factor of $15$ ).

This is true in general: If $a$ is a factor of $b$ , then $b$ is divisible by $a$ , and vice versa.

## Factors and divisibility in polynomials

This knowledge can be applied to polynomials as well.

When two or more polynomials are multiplied, we call each of these polynomials

**factors**of the product.For example, we know that $2x(x+3)=2{x}^{2}+6x$ .
This means that $2x$ and $x+3$ are factors of $2{x}^{2}+6x$ .

Also, one polynomial is

**divisible**by another polynomial if the quotient is also a polynomial.For example, since $\frac{6{x}^{2}}{3x}}=2x$ and since $\frac{6{x}^{2}}{2x}}=3x$ , then $6{x}^{2}$ is divisible by $3x$ and $2x$ . However, since $\frac{4x}{2{x}^{2}}}={\displaystyle \frac{2}{x}$ , we know that $4x$ is $2{x}^{2}$ .

*not divisible*byWith polynomials, we can note the same relationship between factors and divisibility as with integers.

In general, if $p=q\cdot r$ for polynomials $p$ , $q$ , and $r$ , then we know the following:

and$q$ are factors of$r$ .$p$ is divisible by$p$ and$q$ .$r$

### Check your understanding

## Determining factors and divisibility

### Example 1: Is $24{x}^{4}$ divisible by $8{x}^{3}$ ?

To answer this question, we can find and simplify $\frac{24{x}^{4}}{8{x}^{3}}$ . If the result is a monomial, then $24{x}^{4}$ is divisible by $8{x}^{3}$ . If the result is not a monomial, then $24{x}^{4}$ is not divisible by $8{x}^{3}$ .

Since the result is a monomial, we know that $24{x}^{4}$ is divisible by $8{x}^{3}$ . (This also implies that $8{x}^{3}$ is a factor of $24{x}^{4}$ .)

### Example 2: Is $4{x}^{6}$ a factor of $32{x}^{3}$ ?

If $4{x}^{6}$ is a factor of $32{x}^{3}$ , then $32{x}^{3}$ is divisible by $4{x}^{6}$ . So let's find and simplify $\frac{32{x}^{3}}{4{x}^{6}}$ .

Notice that the term $\frac{8}{{x}^{3}}$ is $4{x}^{6}$ $32{x}^{3}$ .

*not*a monomial since it is a quotient, not a product. Therefore we can conclude that*is not*a factor of### A summary

In general, to determine whether one polynomial $p$ is divisible by another polynomial $q$ , or equivalently whether $q$ is a factor of $p$ , we can find and examine $\frac{p(x)}{q(x)}$ .

If the simplified form is a polynomial, then $p$ is divisible by $q$ and $q$ is a factor of $p$ .

### Check your understanding

## Challenge problems

## Why are we interested in factoring polynomials?

Just as factoring integers turned out to be very useful for a variety of applications, so is polynomial factorization!

Specifically, polynomial factorization is very useful in solving quadratic equations and simplifying rational expressions.

If you'd like to see this, check out the following articles:

## What's next?

The next step in the factoring process involves learning how to factor monomials. You can learn about this in our next article.

## Want to join the conversation?

- what is a factor(17 votes)
- A number that multiplies to another number.(49 votes)

- The lesson was a little hard man...(30 votes)
- I'm glad I'm not the only one :/ but if we practice a lot we'll get it eventually!(25 votes)

- Hey all. Confused on last question. Wouldn’t x2+5x be a factor since you just cancel the x2+5x on top and bottom? Or is it not just because of the problem being about Area?(7 votes)
- I realize this was posted 8 months ago, but this is a common mistake so I would like to address it. x^2+5x is not a factor of this expression because it is being added to 4. If that sum were multiplied by 4 instead of added to it, then it would be a factor. The fact that the expression is a sum of x^2+5 and 4 and not a product of the two means that x^2+5 cannot be a factor of x^2+5x+4. I hope that makes sense and clears this up for anyone else wondering the same thing.(41 votes)

- How does this work?

I mean I get it but it's really hard.

(btw im doing 8th grade math and im in 7th grade.)(8 votes)- (Assuming the denominator is a monomial)

First when you are dividing a polynomial, it's better to take each element separated by + or - in the numerator.

For example,

(2x^2 + 6x)/2x

-> 2x^2/2x + 6x/2x

This is basically separating a fraction into smaller fractions.

Then the next thing you will do is think as each variable / constant that is being multiplied not placed together.

Consider 2x^2/2x

So you can first take 2/2 and compute it, which results to 1.

Then you consider x^2/x, which is x^(2-1) = x.

Consider 15x^2y^6 / 10x^4y^3.

First take 15 / 10, which is 3 / 2. Then consider x^2 / x^4, which is 1 / x^2. Last y^6 / y^3 which is y^3.

The reason why this works is because of the fundamental of algebra.

Hope this helps(4 votes)

- What is monomials?(2 votes)
- A monomial is a math statement with exactly one term. A term can be anything with multiplication or division and it can even contain variables, but it cannot contain addition or subtraction. Here are some examples of what is and isn't a monomial.
**IS**a monomial

- 9x

- 14a/2

- 4

- 17ab^2**IS NOT**a monomial

- 14x + 4

- 919 - 7

- x^2 + 4x + 9

An easy way to remember what a monomial is can be to break it down into its Latin/Greek roots.

Mono - nomial**mono**means one**nomial**means number or term

Hope this helps!(16 votes)

- i really don't understand this topic how do you do this(8 votes)
- What's the easiest way to tell a number is a factor of another?(5 votes)
- Think about what 2 numbers
**multiply**together to make that number. For example**3 x 4 = 12**therefore**3 and 4**are**factors**.(6 votes)

- This lesson was easy for me how about y`all?(6 votes)
- the last one got me confused(6 votes)
- i don't get it..?(5 votes)