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# Dividing rational expressions

Learn how to find the quotient of two rational expressions.

#### 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 includes all real numbers except those that make its denominator equal to zero.
We can multiply rational expressions in much the same way that we multiply numerical fractions — by factoring, canceling common factors, and multiplying across.
If this is not familiar to you, you'll want to check out the following articles first:

#### What you will learn in this lesson

In this lesson, you will learn how to divide rational expressions.

## Dividing fractions

To divide two numerical fractions, we multiply the dividend (the first fraction) by the reciprocal of the divisor (the second fraction). For example:
\begin{aligned} &\phantom{=}\dfrac{2}{9}\div{\dfrac{8}{3}}\\\\\\ &=\dfrac{2}{9}\cdot {\dfrac{3}{8}}&&\small{\gray{\text{Multiply by the reciprocal}}}\\ \\ &=\dfrac{\blueD2}{\greenD3\cdot 3}\cdot \dfrac{\greenD3}{\blueD2\cdot 4}&&\small{\gray{\text{Factor numerators & denominators}}}\\\\ &=\dfrac{\blueD{\cancel{2}}}{\greenD{\cancel{3}}\cdot 3}\cdot \dfrac{\greenD{\cancel{3}}}{\blueD{\cancel{2}}\cdot 4}&&\small{\gray{\text{Cancel common factors}}}\\\\ &=\dfrac{1}{12}&&\small{\gray{\text{Multiply across}}} \end{aligned}
We can also use this method to divide rational expressions.

## Example 1: $\dfrac{3x^4}{4}\div\dfrac{9x}{10}$start fraction, 3, x, start superscript, 4, end superscript, divided by, 4, end fraction, divided by, start fraction, 9, x, divided by, 10, end fraction

\begin{aligned} &\phantom{=}\dfrac{3x^4}{4}\div\dfrac{9x}{10}\\\\\\ &=\dfrac{3x^4}{4}\cdot \dfrac{10}{9x}&&\small{\gray{\text{Multiply by the reciprocal}}}\\ \\ &=\dfrac{\blueD3\cdot \greenD{x}\cdot x^3}{\goldD2\cdot 2}\cdot \dfrac{\goldD 2\cdot 5}{\blueD3\cdot 3\cdot \greenD{x}}&&\small{\gray{\text{Factor numerators & denominators}}}\\\\ &=\dfrac{\blueD{\cancel{3}}\cdot \greenD{\cancel{x}}\cdot x^3}{\goldD{\cancel{2}}\cdot 2}\cdot \dfrac{\goldD{\cancel{2}}\cdot 5}{\blueD{\cancel{3}}\cdot 3\cdot \greenD{\cancel{x}}}&&\small{\gray{\text{Cancel common factors}}}\\\\ &=\dfrac{5x^3}{6}&&\small{\gray{\text{Multiply across}}} \end{aligned}
As always, we need to think about restricted values. When dividing two rational expressions, the quotient is undefined...
• for any value that makes either of the original rational expressions undefined,
• and for any value that makes the divisor equal to zero.
To summarize, the expression that is the result of start fraction, A, divided by, B, end fraction, divided by, start fraction, C, divided by, D, end fraction is undefined when either B, equals, 0, C, equals, 0, or D, equals, 0.
Let's examine the dividend and the divisor in this problem to determine any domain restrictions.
• The dividend start fraction, 3, x, start superscript, 4, end superscript, divided by, 4, end fraction is defined for all x-values.
• The divisor start fraction, 9, x, divided by, 10, end fraction is defined for all x-values, and is equal to zero for x, equals, 0.
Therefore, we can conclude that the resulting quotient is defined for x, does not equal, 0. This is our final answer:
start fraction, 5, x, cubed, divided by, 6, end fraction for x, does not equal, 0

1) Divide and simplify the result.
start fraction, 3, divided by, 10, x, squared, end fraction, divided by, start fraction, 6, divided by, 15, x, start superscript, 5, end superscript, end fraction, equals
for x, does not equal

## Example 2: $\dfrac{x^2+x-6}{x^2+3x-10}\div \dfrac{x+3}{x-5}$start fraction, x, squared, plus, x, minus, 6, divided by, x, squared, plus, 3, x, minus, 10, end fraction, divided by, start fraction, x, plus, 3, divided by, x, minus, 5, end fraction

As always, we multiply the dividend by the reciprocal of the divisor. Then we factor, cancel common factors, and multiply across. Finally, we consider restricted values.
\begin{aligned} &\phantom{=}\dfrac{x^2+x-6}{x^2+3x-10}\div \dfrac{x+3}{x-5}\\\\\\ &=\dfrac{x^2+x-6}{x^2+3x-10}\cdot \dfrac{x-5}{x+3}&&\small{\gray{\text{Multiply by the reciprocal}}}\\ \\ &=\dfrac{\blueD{(x+3)}\greenD{(x-2)}}{(x+5)\greenD{(x-2)}}\cdot \dfrac{x-5}{\blueD{x+3}}&&\small{\gray{\text{Factor}}}\\\\ &=\dfrac{\blueD{\cancel{(x+3)}}\greenD{\cancel{(x-2)}}}{(x+5)\greenD{\cancel{(x-2)}}}\cdot \dfrac{(x-5)}{\blueD{\cancel{x+3}}}&&\small{\gray{\text{Cancel common factors}}}\\\\ &=\dfrac{x-5}{x+5}&&\small{\gray{\text{Multiply across}}} \end{aligned}
Let's examine the dividend and the divisor in this problem to determine any domain restrictions. It is easiest to use the factored form of these expressions.
• The dividend start fraction, left parenthesis, x, plus, 3, right parenthesis, left parenthesis, x, minus, 2, right parenthesis, divided by, left parenthesis, x, plus, 5, right parenthesis, left parenthesis, x, minus, 2, right parenthesis, end fraction is defined for x, does not equal, minus, 5, comma, 2.
• The divisor start fraction, x, plus, 3, divided by, x, minus, 5, end fraction is defined for x, does not equal, 5, and is equal to zero for x, equals, minus, 3.
Therefore, we can conclude that the resulting quotient is defined for x, does not equal, minus, 5, comma, minus, 3, comma, 2, comma, 5.
Because of this, we must note that x, does not equal, 5, comma, 2, comma, minus, 3. We do not need to note that x, does not equal, minus, 5, since this is understood from the expression. This is our final answer:
start fraction, x, minus, 5, divided by, x, plus, 5, end fraction for x, does not equal, 5, comma, 2, comma, minus, 3

2) Divide and simplify the result.
start fraction, x, minus, 7, divided by, x, squared, minus, 4, end fraction, divided by, start fraction, x, squared, minus, 6, x, minus, 7, divided by, 2, x, plus, 4, end fraction, equals
What are the restrictions on the domain of the resulting expression?

3) Divide and simplify the result.
start fraction, x, plus, 4, divided by, x, squared, minus, 9, end fraction, divided by, start fraction, x, minus, 1, divided by, x, squared, minus, 4, x, plus, 3, end fraction, equals
What are the restrictions on the domain of the resulting expression?

## Want to join the conversation?

• Example 2: When it is Fully worked back through it is 2/ x^2-x-2, is it appropriate to cancel out the Numerator 2 with the Denominator -2 resulting in - 1/x^2-x to simplify as far as possible?
• No, you can't cancel the 2 in the numerator with the -2 in the denominator.
When you reduce fractions, we cancel out common factors. Factors are items being multiplied.
For example: 15/25 = (3*5)/(5*5). We cancel out a common factor of 5 to get 3/5
If we have: (3+5)/(5+5), we can't cancel the 5's. They are not factors; they are terms (items being added/subtracted).
If you simplify the fraction: (3+5)/(5+5) = 8/10 = 4/5. The common factor is "2", not "5".

So, let's look at your example. The "-2" in the denominator is a term, not a factor. It is being added to the "x^2 - x". Thus, we can't cancel the 2 in the numerator with the "-2" in the denominator.

Hope this helps.
• In the above review it states,
“To summarize, the expression that is the result of A/B divided by C/D is undefined when either B=0, C=0, or D=0.”

Is the result of the division really undefined when D=0?

Perhaps that’s been formally established somewhere, but I’m not really convinced it holds. Help me out on this?

Example:
(x+1)/(x+2) divided by (x+3)/(x-4) … while the divisor might be undefined at x=4, is the quotient, really?

My supposition:
Consider 2 rational numbers, p1/q1 and q2/p2 , where p1, q1, p2, q2 are integers, q1≠0 and q2≠0, then:
p1/q1 divided by q2/p2 is defined for p2=0.

Proposed proof:
Assume the contrary: p1/q1 divided by q2/p2 is not defined for p2=0.

Now given any 2 rational numbers in the form p1/q1 and p2/q2, their product is defined as p1p2/(q1q2) for integers p1, q1, p2, q2 except when q1≠0 or q2≠0.

Furthermore, the division of any 2 rational numbers, p1/q1 divided by q2/p2 is defined as the product p1/q1 and p2/q2 = p1p2/(q1q2), although the product is undefined whenever q1≠0 or q2≠0.

This leads to a contradiction from the assumption above, since the product of 2 rational numbers in the form p1/q1 and p2/q2 does not exclude zero values for p1 and p2.

It follows that the assumption is invalid, and p1/q1 divided by q2/p2 is defined for p2=0. Extension of this argument to the ratio of real numbers and their functions however, appears a bit trickier.

The reason for looking at this (besides getting the correct answer on a test) is that if one has non-zero p1, q1, and q2, and then one also excludes p2=0 upon division of 2 rational numbers in the form p1/q1 and q2/p2, one will never have a zero quotient!
• I used LaTeX to make the notation clearer. Here is the link to my answer:
https://latex.artofproblemsolving.com/miscpdf/flcazhyq.pdf?t=1501862231942
You may have to reload this link after you open it to refresh it since I found a couple spelling errors after posting this answer which I later corrected.
• Why in example two, under "I need help", does it say that the divisor is defined for x not = to -2, and equal to zero for x=7, -1
and in example three, under "I need help", does it say that the divisor is defined for x not = to 3,1 ?
I got the questions correct, but I don't know what the difference is. How does one know when to use which wording?
• It says x not = to -2, and equal to zero for x=7, -1 because 2(x+2) is the denominator of the simplified divisor, and in fractions denominator=0 is undefined, therefore if x=2 the divisor is undefined, so the equation is defined when x not =2.*

the equation=0 for x=7,-1 means (x-7)(x+1)=0. remember our divisor was *multiplied by the reciprocal
. so (x-7)(x+1) was originally in the numerator but when flipped around was in the denominator, therefore x=7 and x=-1 when (x-7)(x+1) is in the numerator of the original divisor. wording being: equal to zero for x=7, -1. but because of being flipped around (multiplied by the reciprocal) x cannot = 7 or -1 or expression would be undefined so your final answer would be

x not = to 7,-1, -2 (only for the divisor).

the wording does not really matter for the actual answer, but in summary equal to zero for x= blank just means numerator=0 of the divisor, since later it gets multiplied by the reciprocal and becomes the denominator.
(1 vote)
• For problem three why is the four left out of the factoring part of the equation?
• Which "4" are you referring to?
The binomial: x+4 can not be factored. So, it doesn't change.
The trinomial: x^2 - 4x + 3
-- We find factors of 3 that add to -4. These are -1 and -3.
-- This creates the factors (x-1)(x-3)
-- You don't see the "-4x" in factored form because it gets created by multiplying the factors.

Hope this helps.
• I notice that now you need to include all numbers causing division by zero, INCLUDING those that were cancelled out. In a previous session you demanded that you exclude those that were cancelled out. Confusing! how is one to know what is expected when you change the expectations/rules.
(1 vote)
• I don't know which "previous session" you refer to. All numbers causing division by zero have to be excluded from the domain, but it is only necessary to note specifically those values that have been cancelled out AND no longer remain in the denominator.
• Example # 3: (x+4)/(x^2-9) / (x-1)/(x^2-4x+3) This has been explained to me once and I thought I understood the issue but don't.

The problem factors to:
(x+4)/(x+3)(x-3)*(x-1)(x-3)/(x-1). The resulting problem factors to x+4/x+3; I got that part correct and fully understand the factoring portion of the problem.

But part of the define portion I don't understand.
I understand that x cannot equal 1,3 but I thought since the denominator of the final solution was x+3, x cannot equal -3 did not have to be listed because it was understood. Plus, why doesn't -4 not have to be listed in the nominator of the problem since -4 makes the nominator equal to zero (0).

Thanks very much. I appreciate the help, this is a great site for math.
• I think this is a case where you just need to pay close attention to the instructions. It asked for all exclusions, so you should include the x not = -3.
Hope this helps.
• in the last problem: The divisor \dfrac{x-1}{(x-3)(x-1)}
​(x−3)(x−1)

​x−1
​​ start fraction, x, minus, 1, divided by, left parenthesis, x, minus, 3, right parenthesis, left parenthesis, x, minus, 1, right parenthesis, end fraction is defined for x\neq 3,1x≠3,1x, does not equal, 3, comma, 1.
Therefore, we can conclude that the resulting quotient is defined for x\neq -3,1,3x≠−3,1,3x, does not equal, minus, 3, comma, 1, comma, 3.

shouldnt it be correct to put just 3 and 1 excluded from de domain? as the expresion (x+4)/x+3 gives explicitly that -3 cant be part of the dom
• Hello can you help me to solve x^2+2x-3 ÷ ×^2+4×-5
X^2-6×+8 ÷ ×^2+3×-10
(1 vote)
• Cube root of 2xsquare divided by the square root of 6x
• "To divide two numerical fractions, we multiply the dividend (the first fraction) by the reciprocal of the divisor (the second fraction)." Why and how does this work? That is, what's the math behind the validity of this rule?
(1 vote)
• Alright, good question.

Let's first try a simple example.
1 ÷ 2 = 1/2
Right?
Isn't it the same thing as
1 x 1/2 = 1/2 ?
As you see, 2 is the reciprocal of 1/2

Another example:
2 ÷ 3 = 2/3
again, isn't it the same thing as
2 x 1/3 = 2/3 ?
And again, 3 is the reciprocal of 1/3.

So now we know that a ÷ b = a x 1/b
Another example, let's say we want to figure out this:
2 ÷ 3/5 = 10/3
since a ÷ b = a x 1/b,
we can say
2 x 1/(3/5) = 10/3
1 over 3 is 1/3, and 1 over 1/5 is 5, so:
2 x 1/3 x 5 = 10/3
we multiply the 1/3 and 5 to get:
2 x 5/3 = 10/3
And if you have noticed, 5/3 is the reciprocal of 3/5.