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# Multiplying rational expressions

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:
\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: $\dfrac{3x^2}{2}\cdot \dfrac{2}{9x}$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.
\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

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

## Example 2: $\dfrac{x^2-x-6}{5x+5}\cdot\dfrac {5}{x-3}$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.
\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.

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?

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?

### What's next?

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

## Want to join the conversation?

• on question 2 if you cancel out the whole denominator won't that just make it 0 and make the the answer /no solution ?
• No. If you cancel out the denominator then it would become 1. Thus, the final answer doesn't have a denominator since it is implied that it's denominator is 1.
• For example 2, the answer x +2/x+1 has two Xs. Can't those Xs be cancelled?
• 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)
• 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?
• 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.
• 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?
• The denominator is: x^2 - 2x - 8. Find factors of -8 that add to the middle term (-2). The factors are -4 and 2.
That creates the binomial factors of (x-4)(x+2).

The instructions said to select all that apply. x not = -2 applies, as does x not = 4 and 3
• 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 ?
• A 0 in the denominator (in your example: (x+2)/0 ) is undefined, but a numerator in the denominator (in your example: (0/x+1)) isn't undefined. 0/x is 0.
• 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)
• Because the instructions say to.... Follow the instructions. It asked you to select ALL that apply.
• Find the numerical value for
Start Fraction x minus 9 Over 9 End Fraction
x−9/9 when x=13
• Before simplifying the multiplying of fractions, you can divide out the common factor. true or false
• So im very confused on how to prove whether it's undefined?? What exactly do you do?
• Take just the denominator, then set it equal to zero. Solve that for x. That's where it's undefined. For example:

Y=(bla bla bla)/(x-3)

Is undefined when x-3=0, or in other words when x-3
• 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.