If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Get ready for AP® Calculus

### Unit 1: Lesson 19

Reducing rational expressions to lowest terms

# Reducing rational expressions to lowest terms

Learn what it means to reduce a rational expression to lowest terms, and how it's done!

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

A rational expression is a ratio of two polynomials. The domain of a rational expression is all real numbers except those that make the denominator equal to zero.
For example, the domain of the rational expression start fraction, x, plus, 2, divided by, x, plus, 1, end fraction is all real numbers except start text, negative, 1, end text, or x, does not equal, minus, 1.
If this is new to you, we recommend that you check out our intro to rational expressions.
You should also know how to factor polynomials for this lesson.

#### What you will learn in this lesson

In this article, we will learn how to reduce rational expressions to lowest terms by looking at several examples.

## Introduction

A rational expression is reduced to lowest terms if the numerator and denominator have no factors in common.
We can reduce rational expressions to lowest terms in much the same way as we reduce numerical fractions to lowest terms.
For example, start fraction, 6, divided by, 8, end fraction reduced to lowest terms is start fraction, 3, divided by, 4, end fraction. Notice how we canceled a common factor of 2 from the numerator and the denominator:
\begin{aligned} \dfrac68&= \dfrac{2\cdot 3}{2\cdot 4} \\\\ &= \dfrac{\tealE{\cancel{2}}\cdot 3}{\tealE{\cancel{2}}\cdot 4} \\\\ &= \dfrac{3}{4} \end{aligned}

## Example 1: Reducing $\dfrac{x^2+3x}{x^2+5x}$start fraction, x, squared, plus, 3, x, divided by, x, squared, plus, 5, x, end fraction to lowest terms

Step 1: Factor the numerator and denominator
The only way to see if the numerator and denominator share common factors is to factor them!
start fraction, x, squared, plus, 3, x, divided by, x, squared, plus, 5, x, end fraction, equals, start fraction, x, left parenthesis, x, plus, 3, right parenthesis, divided by, x, left parenthesis, x, plus, 5, right parenthesis, end fraction
Step 2: List restricted values
At this point, it is helpful to notice any restrictions on x. These will carry over to the simplified expression.
Since division by 0 is undefined, here we see that start color #0c7f99, x, does not equal, 0, end color #0c7f99 and start color #7854ab, x, does not equal, minus, 5, end color #7854ab.
start fraction, x, left parenthesis, x, plus, 3, right parenthesis, divided by, start color #0c7f99, x, end color #0c7f99, start color #7854ab, left parenthesis, x, plus, 5, right parenthesis, end color #7854ab, end fraction
Step 3: Cancel common factors
Now notice that the numerator and denominator share a common factor of x. This can be canceled out.
\begin{aligned} \dfrac{\tealE x(x+3)}{\tealD x(x+5)}&=\dfrac{\tealE {\cancel {x}}(x+3)}{\tealE{\cancel x}(x+5)} \\\\ &=\dfrac{x+3}{x+5} \end{aligned}
Recall that the original expression is defined for x, does not equal, 0, comma, minus, 5. The reduced expression must have the same restrictions.
Because of this, we must note that x, does not equal, 0. We do not need to note that x, does not equal, minus, 5, since this is understood from the expression.
In conclusion, the reduced form is written as follows:
start fraction, x, plus, 3, divided by, x, plus, 5, end fraction for x, does not equal, 0

### A note about equivalent expressions

Original expressionReduced expression
start fraction, x, squared, plus, 3, x, divided by, x, squared, plus, 5, x, end fractionstart fraction, x, plus, 3, divided by, x, plus, 5, end fraction for x, does not equal, 0
The two expressions above are equivalent. This means their outputs are the same for all possible x-values. The table below illustrates this for x, equals, 2.
Original expressionReduced expression
Evaluation at start color #7854ab, x, equals, 2, end color #7854ab\begin{aligned}\dfrac{(\purpleD{2})^2+3(\purpleD{2})}{(\purpleD{2})^2+5(\purpleD{2})}&=\dfrac{10}{14}\\\\&=\dfrac{\purpleD{{2}}\cdot 5}{\purpleD{{2}}\cdot 7}\\\\&=\dfrac{\purpleD{\cancel{2}}\cdot 5}{\purpleD{\cancel{2}}\cdot 7}\\\\&=\dfrac{5}{7}\end{aligned}\begin{aligned}\dfrac{\purpleD{2}+3}{\purpleD{2}+5}&=\dfrac{5}{7}\\\\&\phantom{=\dfrac57}\\\\&\phantom{=\dfrac57}\\\\&\phantom{=\dfrac57}\end{aligned}
NoteThe result is reduced to lowest terms by canceling a common factor of start color #7854ab, 2, end color #7854ab.The result is already reduced to lowest terms because the factor of x (in this case start color #7854ab, x, equals, 2, end color #7854ab), was already canceled when we reduced the expression to lowest terms.
For this reason, the two expressions have the same value for the same input. However, values that make the original expression undefined often break this rule. Notice how this is the case with start color #7854ab, x, equals, 0, end color #7854ab.
Original expressionReduced expression (without restriction)
Evaluation at start color #7854ab, x, equals, 0, end color #7854ab\begin{aligned}\dfrac{(\purpleD{0})^2+3(\purpleD{0})}{(\purpleD{0})^2+5(\purpleD{0})}&=\dfrac{0}{0}\\\\&=\text{undefined}\end{aligned}\begin{aligned}\dfrac{\purpleD{0}+3}{\purpleD{0}+5}&=\dfrac{3}{5}\\\\&\phantom{\text{undefined}}\end{aligned}
Because the two expressions must be equivalent for all possible inputs, we must require x, does not equal, 0 for the reduced expression.

Note that we cannot cancel the x's in the expression below. This is because these are terms rather than factors of the polynomials!
start fraction, x, plus, 3, divided by, x, plus, 5, end fraction, does not equal, start fraction, 3, divided by, 5, end fraction
This becomes clear when looking at a numerical example. For example, suppose start color #7854ab, x, equals, 2, end color #7854ab.
start fraction, start color #7854ab, 2, end color #7854ab, plus, 3, divided by, start color #7854ab, 2, end color #7854ab, plus, 5, end fraction, does not equal, start fraction, 3, divided by, 5, end fraction
As a rule, we can only cancel if the numerator and denominator are in factored form!

### Summary of the process for reducing to lowest terms

• Step 1: Factor the numerator and the denominator.
• Step 2: List restricted values.
• Step 3: Cancel common factors.
• Step 4: Reduce to lowest terms and note any restricted values not implied by the expression.

Problem 1
Reduce start fraction, 6, x, plus, 20, divided by, 2, x, plus, 10, end fraction to lowest terms.

Problem 2
Reduce start fraction, x, cubed, minus, 3, x, squared, divided by, 4, x, squared, minus, 5, x, end fraction to lowest terms.
for x, does not equal

## Example 2: Reducing $\dfrac{x^2-9}{x^2+5x+6}$start fraction, x, squared, minus, 9, divided by, x, squared, plus, 5, x, plus, 6, end fraction to lowest terms

Step 1: Factor the numerator and denominator
start fraction, x, squared, minus, 9, divided by, x, squared, plus, 5, x, plus, 6, end fraction, equals, start fraction, left parenthesis, x, minus, 3, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, divided by, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, end fraction
Step 2: List restricted values
Since division by 0 is undefined, here we see that start color #0c7f99, x, does not equal, minus, 2, end color #0c7f99 and start color #7854ab, x, does not equal, minus, 3, end color #7854ab.
start fraction, left parenthesis, x, minus, 3, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, divided by, start color #0c7f99, left parenthesis, x, plus, 2, right parenthesis, end color #0c7f99, start color #7854ab, left parenthesis, x, plus, 3, right parenthesis, end color #7854ab, end fraction
Step 3: Cancel common factors
Notice that the numerator and denominator share a common factor of start color #208170, x, plus, 3, end color #208170. This can be canceled out.
\begin{aligned} \dfrac{(x-3)\tealE{(x+3)}}{(x+2)\tealE{(x+3)}}&=\dfrac{(x-3)\tealE{\cancel{(x+3)}}}{(x+2)\tealE{\cancel{(x+3)}}} \\\\ &=\dfrac{x-3}{x+2} \end{aligned}
We write the reduced form as follows:
start fraction, x, minus, 3, divided by, x, plus, 2, end fraction for x, does not equal, minus, 3
The original expression requires x, does not equal, minus, 2, comma, minus, 3. We do not need to note that x, does not equal, minus, 2, since this is understood from the expression.

### Check for understanding

Problem 3
Reduce start fraction, x, squared, minus, 3, x, plus, 2, divided by, x, squared, minus, 1, end fraction to lowest terms.

Problem 4
Reduce start fraction, x, squared, minus, 2, x, minus, 15, divided by, x, squared, plus, x, minus, 6, end fraction to lowest terms.
for x, does not equal

## Want to join the conversation?

• I don't understand much the restricted values. Can someone help me to understand it please?
• If we have an equation (x^2 + 3x) / (x^2 + 5x) , we can simplify it to (x+3) / (x+5) . At first glance, the two equations seem to be equal, but they are actually not!
Remember that x can be any value - let just try the values -1, 1 and 0 for x in the two equations above.
For x = -1,
(x^2 + 3x) / (x^2 + 5x) = (1 - 3) / (1 - 5) = -2/-4 = 0.5
(x+3) / (x+5) = (-1 + 3)/ (-1 + 5) = 2/4 = 0.5
Both equations are equivalent as both of them give the value 0.5
For x = 1,
(x^2 + 3x) / (x^2 + 5x) = (1 + 3) / (1 + 5) = 4/6 = 2/3
(x+3) / (x+5) = (1 + 3)/ (1 + 5) = 4/6 = 2/3
Both equations are equivalent as both of them give the value 2/3
For x = 0,
(x^2 + 3x) / (x^2 + 5x) = (0 + 0) / (0 + 0) = undefined !
(x+3) / (x+5) = (0 + 3)/ (0 + 5) = 3/5 !
Both equations are NOT equivalent as both of them give different values in the case of x =0
So from the above, whenever the denominator results in a value of 0, we get an undefined value for the expression. Therefore we need to put the restriction x not equal to 0 in this case on the equation (x+3) / (x+5). With this restriction, the simplified equation is now equivalent to the original rational expression.
• How do I remember all of these steps?
• Practice, practice, practice. Eventually, you will do it enough that you don’t even think about it. You could come up with an acronym to help you, but the final goal is that you won’t need it.
• In #3, why can't x be equal to 1? I understand why it can't equal -1, but I don't see where 1 causes the expression to be undefined.
• Factor both the numerator and the denominator.
Numerator = (x-2)(x-1)
Denominator = (x-1)(x+1)
The factor of (x-1) in the denominator would cause the an undefined if x=1
Even though this factor cancels out, you need to maintain the restriction that x can't = 1 to be consistent with the original expression.
Hope this helps.
• This is very confusing? How will this help us in real life?
• It probably, most likely, won't. although some parts of the business and trade use this math, although I don't know any, there might be some.
• what makes an expression undefined?
• With rational expressions (fractions), they become undefined if the denominator = 0. If a denominator = 0, then we are dividing by 0, which is undefined.
• Just to get this clear, the way you figure out restricted values is like this, right?:
Let's say you have and equation that you simplified to q+1/q+2.
The restriction is that q cannot be equal to -2, correct?
I'm studying for SSATs, so I need to make sure I know this stuff. It's always been a little confusing, but now I think I've got it. Just need this clarified. Also, when in the article it said that "We do not need to note that x≠−5, since this is understood from the expression." Why don't we need to note that? How is it understood? Why wouldn't you list all the restrictions, instead of just one?
• Let's use the expression:

(x+1)(x+2) divided by (x+3)(x+2)

This expression is undefined at x=-3 and x=-2 because both values will make the denominator equal to 0. Division by 0 is undefined. You can factor out (x+2) from both the numerator and the denominator, resulting in (x+1) divided by (x+3).

Try plugging in x=-2 into both expressions. Even though the original expression was undefined for x=-2, the simplified expression gets a different answer (-1) for that same input. Therefore, the expressions are not truly equal unless we add the restriction that x cannot equal -2.

Now try x=-3. Notice that both expressions are undefined for x=-3. Therefore, the expressions are still equivalent. You can also say that the simplified expression "implies" that x=-3 is undefined because the simplified expression is undefined for x=-3. Therefore it isn't required to remind everyone that x cannot equal -3 because you can figure that out with just the simplified expression. You can state as such if you want. On the other hand, you have to state x cannot equal -2 because that is not "implied" in the simplified expression.

I hope that helped. I know it was a lot to take in...
• hmmm it's a bit weird, but ok