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## Algebra 1

### Unit 13: Lesson 4

Introduction to factoring

# 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 monomial is an expression that is the product of constants and nonnegative integer powers of x, like 3, x, squared. A polynomial is an expression that consists of a sum of monomials, like 3, x, squared, plus, 6, x, minus, 1.

## 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, equals, 2, dot, 7, we know that 2 and 7 are factors of 14.
One number is divisible by another number if the result of the division is an integer.
For example, since start fraction, 15, divided by, 3, end fraction, equals, 5 and start fraction, 15, divided by, 5, end fraction, equals, 3, then 15 is divisible by 3 and 5. However, since start fraction, 9, divided by, 4, end fraction, equals, 2, point, 25, then 9 is not divisible by 4.
Notice the mutual relationship between factors and divisibility:
Since start color #e07d10, 14, end color #e07d10, equals, start color #11accd, 2, end color #11accd, dot, 7 (which means 2 is a factor of 14), we know that start fraction, start color #e07d10, 14, end color #e07d10, divided by, start color #11accd, 2, end color #11accd, end fraction, equals, 7 (which means 14 is divisible by 2).
start underbrace, start color #a75a05, 14, end color #a75a05, equals, start color #0c7f99, 2, end color #0c7f99, dot, 7, end underbrace, start subscript, 2, start text, space, i, s, space, a, space, f, a, c, t, o, r, space, o, f, space, end text, 14, end subscript, \longrightarrow, start underbrace, start fraction, start color #a75a05, 14, end color #a75a05, divided by, start color #0c7f99, 2, end color #0c7f99, end fraction, equals, 7, end underbrace, start subscript, 14, start text, space, i, s, space, d, i, v, i, s, i, b, l, e, space, b, y, space, end text, 2, end subscript
In the other direction, since start fraction, start color #e07d10, 15, end color #e07d10, divided by, start color #11accd, 3, end color #11accd, end fraction, equals, 5 (which means 15 is divisible by 3), we know that start color #e07d10, 15, end color #e07d10, equals, start color #11accd, 3, end color #11accd, dot, 5 (which means 3 is a factor of 15).
start underbrace, start fraction, start color #a75a05, 15, end color #a75a05, divided by, start color #0c7f99, 3, end color #0c7f99, end fraction, equals, 5, end underbrace, start subscript, 15, start text, space, i, s, space, d, i, v, i, s, i, b, l, e, space, b, y, space, end text, 3, end subscript, \longrightarrow, start underbrace, start color #a75a05, 15, end color #a75a05, equals, start color #0c7f99, 3, end color #0c7f99, dot, 5, end underbrace, start subscript, 3, start text, space, i, s, space, a, space, f, a, c, t, o, r, space, o, f, space, end text, 15, end subscript
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 2, x, left parenthesis, x, plus, 3, right parenthesis, equals, 2, x, squared, plus, 6, x. This means that 2, x and x, plus, 3 are factors of 2, x, squared, plus, 6, x.
Also, one polynomial is divisible by another polynomial if the quotient is also a polynomial.
For example, since start fraction, 6, x, squared, divided by, 3, x, end fraction, equals, 2, x and since start fraction, 6, x, squared, divided by, 2, x, end fraction, equals, 3, x, then 6, x, squared is divisible by 3, x and 2, x. However, since start fraction, 4, x, divided by, 2, x, squared, end fraction, equals, start fraction, 2, divided by, x, end fraction, we know that 4, x is not divisible by 2, x, squared.
With polynomials, we can note the same relationship between factors and divisibility as with integers.
start underbrace, start color #0c7f99, 2, x, end color #0c7f99, dot, left parenthesis, x, plus, 3, right parenthesis, equals, start color #a75a05, 2, x, squared, plus, 6, x, end color #a75a05, end underbrace, start subscript, 2, x, start text, space, i, s, space, a, space, f, a, c, t, o, r, space, o, f, space, end text, 2, x, squared, plus, 6, x, end subscript, \longrightarrow, start underbrace, start fraction, start color #a75a05, 2, x, squared, plus, 6, x, end color #a75a05, divided by, start color #0c7f99, 2, x, end color #0c7f99, end fraction, equals, x, plus, 3, end underbrace, start subscript, 2, x, squared, plus, 6, x, start text, space, i, s, space, d, i, v, i, s, i, b, l, e, space, b, y, space, end text, 2, x, end subscript
start underbrace, start fraction, start color #a75a05, 6, x, squared, end color #a75a05, divided by, start color #0c7f99, 3, x, end color #0c7f99, end fraction, equals, 2, x, end underbrace, start subscript, 6, x, squared, start text, space, i, s, space, d, i, v, i, s, i, b, l, e, space, b, y, space, end text, 3, x, end subscript, \longrightarrow, start underbrace, start color #0c7f99, 3, x, end color #0c7f99, dot, left parenthesis, 2, x, right parenthesis, equals, start color #a75a05, 6, x, squared, end color #a75a05, end underbrace, start subscript, 3, x, start text, space, i, s, space, a, space, f, a, c, t, o, r, space, o, f, space, end text, 6, x, squared, end subscript
In general, if p, equals, q, dot, r for polynomials p, q, and r, then we know the following:
• q and r are factors of p.
• p is divisible by q and r.

1) Complete the sentence about the relationship expressed by 3, x, left parenthesis, x, plus, 2, right parenthesis, equals, 3, x, squared, plus, 6, x.
x, plus, 2 is
3, x, squared, plus, 6, x, and 3, x, squared, plus, 6, x is
x, plus, 2.

2) A teacher writes the following product on the board:
left parenthesis, 3, x, squared, right parenthesis, left parenthesis, 4, x, right parenthesis, equals, 12, x, cubed
Miles concludes that 3, x, squared is a factor of 12, x, cubed.
Jude concludes that 12, x, cubed is divisible by 4, x.
Who is correct?

## Determining factors and divisibility

### Example 1: Is $24x^4$24, x, start superscript, 4, end superscript divisible by $8x^3$8, x, cubed?

To answer this question, we can find and simplify start fraction, 24, x, start superscript, 4, end superscript, divided by, 8, x, cubed, end fraction. If the result is a monomial, then 24, x, start superscript, 4, end superscript is divisible by 8, x, cubed. If the result is not a monomial, then 24, x, start superscript, 4, end superscript is not divisible by 8, x, cubed.
\begin{aligned}\dfrac{24x^4}{8x^3}&=\dfrac{24}{8}\cdot\dfrac{x^4}{x^3}\\ \\ &=3\cdot x^1&&{\gray{\dfrac{a^m}{a^n}=a^{m-n}}}\\ \\ &=3x \end{aligned}
Since the result is a monomial, we know that 24, x, start superscript, 4, end superscript is divisible by 8, x, cubed. (This also implies that 8, x, cubed is a factor of 24, x, start superscript, 4, end superscript.)

### Example 2: Is $4x^6$4, x, start superscript, 6, end superscript a factor of $32x^3$32, x, cubed?

If 4, x, start superscript, 6, end superscript is a factor of 32, x, cubed, then 32, x, cubed is divisible by 4, x, start superscript, 6, end superscript. So let's find and simplify start fraction, 32, x, cubed, divided by, 4, x, start superscript, 6, end superscript, end fraction.
\begin{aligned}\dfrac{32x^3}{4x^6}&=\dfrac{32}{4}\cdot\dfrac{x^3}{x^6}\\ \\ &=8\cdot x^{-3}&&{\gray{\dfrac{a^m}{a^n}=a^{m-n}}}\\ \\ &=8\cdot \dfrac{1}{x^3}&&{\gray{a^{-m}=\dfrac{1}{a^m}}}\\ \\ &=\dfrac{8}{x^3} \end{aligned}
Notice that the term start fraction, 8, divided by, x, cubed, end fraction is not a monomial since it is a quotient, not a product. Therefore we can conclude that 4, x, start superscript, 6, end superscript is not a factor of 32, x, cubed.

### 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 start fraction, p, left parenthesis, x, right parenthesis, divided by, q, left parenthesis, x, right parenthesis, end fraction.
If the simplified form is a polynomial, then p is divisible by q and q is a factor of p.

3) Is 30, x, start superscript, 4, end superscript divisible by 2, x, squared?

4) Is 12, x, squared a factor of 6, x?

## Challenge problems

5*) Which of the following monomials are factors of 15, x, squared, y, start superscript, 6, end superscript ?

Factor
Not Factor
3, x, squared, y, start superscript, 5, end superscript
5, x
10, x, start superscript, 4, end superscript, y, cubed

6*) The area of a rectangle with height x, plus, 1 units and base x, plus, 4 units is x, squared, plus, 5, x, plus, 4 square units.
An area model for a rectangle that has a height of x plus one and a width of x plus four. The area of the rectangle is x squared plus five x plus four.
Which of the following are factors of x, squared, plus, 5, x, plus, 4?

## 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
• A number that multiplies to another number.
• 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?
• 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.
• The lesson was a little hard man...
• I'm glad I'm not the only one :/ but if we practice a lot we'll get it eventually!
• This lesson was easy for me how about y`all?
• What's the easiest way to tell a number is a factor of another?
• Also 6 and 2 are factors because 6 x 2 = 12. Which make them factors because they can be divided from 12
(1 vote)
• i really don't understand this topic how do you do this
• I dint get numbers 6 and5 I don't see how you got that answer
• combination of these could be a factor, thus 5x and 3x^2 y^5 is okay, 10x^4 y^3 is not because no way to get 10 and no way to get 4 x's
The numbers could only be 3,5, and 15, x could only be to 1st or 2nd, ys up to power 6
6 is straight forward (so how they ask the question must be confusing
(1 vote)
• Does the factor of a polynomial always have to be a monomial?

e.g. 3(3x^2 + 3x + 3)
factors would be 3 and (3x^2 + 3x + 3)?