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.

Main content

Integrated math 3

Course: Integrated math 3 > Unit 13

Lesson 6: Multiplying & dividing rational expressions
Math
Integrated math 3
Rational functions
Multiplying & dividing rational expressions
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

Multiplying rational expressions

Google Classroom
Learn how to find the product 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 simplify rational expressions by canceling common factors in the numerator and the denominator.
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 multiply rational expressions.

Multiplying fractions

To start, let's recall how to multiply numerical fractions.
Consider this example:
=34⋅109=32⋅2⋅2⋅53⋅3Factor numerators and denominators=32⋅2⋅2⋅53⋅3Cancel common factors=56Multiply across\begin{aligned} &\phantom{=}\dfrac{3}{4}\cdot\dfrac{10}{9}\\\\ &=\dfrac{\greenD3}{\blueD2\cdot 2}\cdot \dfrac{\blueD2\cdot 5}{\greenD3\cdot 3} &&\small{\gray{\text{Factor numerators and denominators}}} \\\\ &=\dfrac{\greenD{\cancel{3}}}{\blueD{\cancel{2}}\cdot 2}\cdot \dfrac{\blueD{\cancel{2}}\cdot 5}{\greenD{\cancel{3}}\cdot 3}&&\small{\gray{\text{Cancel common factors}}} \\\\ &=\dfrac{5}{6}&&\small{\gray{\text{Multiply across}}} \end{aligned}
In conclusion, to multiply two numerical fractions, we factored, canceled common factors, and multiplied across.

Example 1: start fraction, 3, x, squared, divided by, 2, end fraction, dot, start fraction, 2, divided by, 9, x, end fraction

We can multiply rational expressions in much the same way as we multiply numerical fractions.
=3x22⋅29x=3⋅x⋅x2⋅23⋅3⋅xFactor numerators and denominators(Note x≠0)=3⋅x⋅x2⋅23⋅3⋅xCancel common factors=x3Multiply across\begin{aligned} &\phantom{=}\dfrac{3x^2}{2}\cdot\dfrac{2}{9x}\\\\\\ &=\dfrac{3\cdot x\cdot x}{2}\cdot \dfrac{2}{3\cdot 3\cdot \goldD x}&& \small{\gray{\text{Factor numerators and denominators}}}\\ \\ &\quad \small{(\text{Note } \goldD{x\neq 0})}\\ \\ \\&=\dfrac{\blueD{\cancel{3}}\cdot \greenD{\cancel{ x}}\cdot x}{\purpleC{\cancel{2}}}\cdot \dfrac{\purpleC{\cancel{2}}}{\blueD{\cancel{ 3}}\cdot 3\cdot \greenD{\cancel{ x}}}&& \small{\gray{\text{Cancel common factors}}} \\ \\ &=\dfrac{x}{3}&&\small{\gray{\text{Multiply across}}} \end{aligned}
Recall that the original expression is defined for x, does not equal, 0. The simplified product must have the same restictions. Because of this, we must note that x, does not equal, 0.
We write the simplified product as follows:
start fraction, x, divided by, 3, end fraction for x, does not equal, 0

Check your understanding

1) Multiply and simplify the result.
start fraction, 4, x, start superscript, 6, end superscript, divided by, 5, end fraction, dot, start fraction, 1, divided by, 12, x, cubed, end fraction, equals
for x, does not equal
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Example 2: start fraction, x, squared, minus, x, minus, 6, divided by, 5, x, plus, 5, end fraction, dot, start fraction, 5, divided by, x, minus, 3, end fraction

Once again, we factor, cancel any common factors, and then multiply across. Finally, we make sure to note all restricted values.
=x2−x−65x+5⋅5x−3=(x−3)(x+2)5⋅(x+1)⋅5x−3Factor(Note x≠−1, and x≠3)=(x−3)(x+2)5⋅(x+1)⋅5x−3Cancel common factors=x+2x+1Multiply across\begin{aligned} &\phantom{=}\dfrac{x^2-x-6}{5x+5}\cdot\dfrac {5}{x-3}\\\\\\ &=\dfrac{(x-3)(x+2)}{5\cdot \goldD{(x+1)}}\cdot \dfrac{5}{\maroonD{x-3}}&&\small{\gray{\text{Factor}}}\\ \\ &\quad \small{(\text{Note }\goldD{x\neq -1}}, \text{ and }\maroonD{x\neq 3} )\\ \\ &=\dfrac{\blueD{\cancel{(x-3)}}{(x+2)}}{\greenD{\cancel{5}}\cdot({x+1})}\cdot \dfrac{{\greenD{\cancel{5}}}}{\blueD{\cancel{x-3}}}&&\small{\gray{\text{Cancel common factors}}}\\ \\ &=\dfrac{x+2}{x+1}&&\small{\gray{\text{Multiply across}}} \end{aligned}
The original expression is defined for x, does not equal, minus, 1, comma, 3. The simplified product must have the same restrictions.
In general, the product of two rational expressions is undefined for any value that makes either of the original rational expressions undefined.

Check your understanding

2) Multiply and simplify the result.
start fraction, 5, x, cubed, divided by, 5, x, plus, 10, end fraction, dot, start fraction, x, squared, minus, 4, divided by, x, squared, end fraction, equals
What are all the restrictions on the domain of the resulting expression?
Choose all answers that apply:

3) Multiply and simplify the result.
start fraction, x, squared, minus, 9, divided by, x, squared, minus, 2, x, minus, 8, end fraction, dot, start fraction, x, minus, 4, divided by, x, minus, 3, end fraction, equals
What are all the restrictions on the domain of the resulting expression?
Choose all answers that apply:

What's next?

If you feel good about your multiplication skills, you can move on to dividing rational expressions.

Want to join the conversation?

Log in
  • leaf green style avatar for user Emma Baumgartel
    Posted 7 years ago. Direct link to Emma Baumgartel's post “on question 2 if you canc...”
    on question 2 if you cancel out the whole denominator won't that just make it 0 and make the the answer /no solution ?
    (6 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Ashley Valdez
    Posted 4 years ago. Direct link to Ashley Valdez's post “For example 2, the answer...”
    For example 2, the answer x +2/x+1 has two Xs. Can't those Xs be cancelled?
    (6 votes)
    Default Khan Academy avatar avatar for user
    • stelly green style avatar for user may lin
      Posted 2 years ago. Direct link to may lin's post “No, because the x is not ...”
      No, because the x is not a factor like say;

      x(x+2)/x(x+4) it's a common factor so the x can be canceled out

      x+2/x+1, you can't cancel out the x because say x = 1

      2+1/1+1 = 3/2

      if you were to cancel out the 1 (or x) you'd get 2/1=2 which is not the same as 3/2.

      this is because it's not something that you factored out from both numerator and denominator, it's something you're adding/subtracting which could change the value.
      (1 vote)
  • marcimus purple style avatar for user Hannah Woods
    Posted 5 years ago. Direct link to Hannah Woods's post “I have absolutely no idea...”
    I have absolutely no idea how you figure out which numbers x cannot equal. I know it says it is the numbers that make the original expression undefined but how do you find that out?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • stelly blue style avatar for user Kim Seidel
      Posted 5 years ago. Direct link to Kim Seidel's post “Rational expressions are ...”
      Rational expressions are fractions. Fractions become undefined if the denominator is = 0.
      For example: 5/0 = undefined.
      Now, if this was 5/x, then it is undefined only when x=0. So x can't = 0.
      If you have an expression of: 5/(x-2), then you look at what would make x-2 = 0. If x=2, this fraction would be undefined.

      When you have a rational expression that you are simplifying, any time you reduce the fraction, you need to ensure that the restrictions associated with the original fraction are maintained.
      For example:
      [(x-2)(x-5)] / [(x-2) (x+7)} would have restrictions of x not equal 2 or -7 because these both cause the denominator to become 0. At this point, we don't have to explicitly state the restrictions because they can be derived from the expression. have to explicitly
      Once reduced (we cancel out the common factor of (x-2)), then we have (x-5) / (x+7). To ensure we maintain the original restrictions, we must explicitly state that "x not equal 2" because this can no longer be derived from the expression.

      Hope this helps.
      (9 votes)
  • duskpin ultimate style avatar for user Apolonio, Morgan
    Posted 6 years ago. Direct link to Apolonio, Morgan's post “For question number 3, ho...”
    For question number 3, how does the denominator of the first expression factor into (x−4)(x+2)? Also for the same number why di they included x=/-2 if you can tell that from the simplified expression?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user N N
    Posted 2 years ago. Direct link to N N's post “For the example 2: Why we...”
    For the example 2:
    Why we don't have to care about numerators? if x = -2 for x+2 / x+1, will it be undefined? Why we don't have to specify x != -2 ?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • orange juice squid orange style avatar for user Chris McKnight
    Posted 7 years ago. Direct link to Chris McKnight's post “For Q3, why do we have to...”
    For Q3, why do we have to specify x does not equal -2 when this is obvious from the simplified form of the expression too? In the videos Sal only gives the values x is undefined for when that value is not clear in the simplified form
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Brandon Kyle Tyson
    Posted 6 years ago. Direct link to Brandon Kyle Tyson's post “Find the numerical value ...”
    Find the numerical value for
    Start Fraction x minus 9 Over 9 End Fraction
    x−9/9 when x=13
    (2 votes)
    Default Khan Academy avatar avatar for user
  • female robot amelia style avatar for user Kathy Mosca
    Posted 3 years ago. Direct link to Kathy Mosca's post “Before simplifying the mu...”
    Before simplifying the multiplying of fractions, you can divide out the common factor. true or false
    (2 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Autumn Rogers
    Posted 6 years ago. Direct link to Autumn Rogers's post “So im very confused on ho...”
    So im very confused on how to prove whether it's undefined?? What exactly do you do?
    (0 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Fred Haynes
    Posted 2 years ago. Direct link to Fred Haynes's post “On problem # 3 why is X c...”
    On problem # 3 why is X cannot equal -2 when x+2 is part of the remaining expression. I've seen it counted wrong both ways in other examples and I am confused as whether to count it or not.

    Thanks in advance.
    (1 vote)
    Default Khan Academy avatar avatar for user
    • duskpin ultimate style avatar for user Andrzej Olsen
      Posted 2 years ago. Direct link to Andrzej Olsen's post “Hello! In this problem, ...”
      Hello!

      In this problem, we're not looking for the simplified expression, we're just looking for the values that make that first expression undefined. If you wanted to, you could just stop at the "factor" step and solve for all the x values that make you divide by 0.

      If they did ask for the simplified expression, then the answer would be
      (x+3)/(x+2), x ≠ 3, 4
      because in this scenario the x ≠ -2 is implied.

      Hope this clears it up.
      (2 votes)