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Definitions and Simplifying
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Multiplying Rational Expressions

Required Material

This section requires you to already understand some techniques for working with polynomials. You can find more information on those techniques on these pages:

Multiplying rational expressions is, more than anything else, an exercise in simplifying. Just like when you're multiplying fractions, the basic procedure is to multiply the numerators and the denominators separately then simplify or reduce the result. As we saw in the section on simplifying rational expressions. When dealing with rational expressions, I've always found it easier to simplify first then multiply things out. The examples below will illustrate why. The entire multiplication procedure goes like this:

1. Factor all of the polynomials in both rational expressions.
2. Cancel any factors that appear in both the numerator and the denominator.
3. Multiply the numerators and the denominators separately to get the final answer.
4. Exclude any values that you canceled form the denominator.

Example 1

Multiply $\frac{x^2+3x+2}{x-5}\cdot\frac{x-5}{x^2-1}$.

The first step in the multiplication process is to factor all of the polynomials. In this case, that will give us

$$\frac{(x+1)(x+2)}{x-5}\cdot\frac{x - 5}{(x+1)(x-1)}$$

I've marked the factors that appear in both the numerator and the denominator in bold and red. Those are the ones that can be canceled. Notice that it doesn't matter which of the two expressions a factor appears in. As long as a factor appears in one of the numerators and one of the denominators we can cancel it. Remember that we can only do this when we're multiplying. If we were adding/subtracting the fractions we wouldn't be able to cancel this way.

If I cancel out the common factors I highlighted, we're left with:

$$\frac{x + 2}{1}\cdot\frac{1}{x - 1}$$

That step illustrates another common point of confusion for students. You can never have "nothing" in the numerator or the denominator. If everything in one of those places gets canceled, we'll put a 1 there to fill the position. Now we can get our final answer by multiplying the numerators and the denominators separately.

$$\frac{(x + 2)\cdot1}{1\cdot(x - 1)}=\frac{x + 2}{x - 1}$$

This brings us to step four, which might have seemed a little confusing. At this point, you might be tempted to say that the original fractions equals our final answer and you would be right with two small exceptions. If you look back at the original expression and think back to the section on undefined values, you'll see that the two original polynomials are undefined at x = -1, 1 and 5. Our final answer and the original expression are equal at every value of x except for -1 and 5. Our simplified version is defined for those values where the original expression isn't. The last step in the multiplication/simplification process is to note that the expression still has to be undefined at x = -1 and 5, i.e. at the points where the factors that we canceled out of the denominator are 0.

Example 2

Multiply $\frac{2x^2+7x-4}{2x^2-9x+4}\cdot\frac{x + 3}{x + 1}$.

We'll start off again by factoring.

$$\frac{(2x - 1)(x + 4)}{(2x - 1)(x - 4)}\cdot\frac{x + 3}{x + 1}$$

I've again marked the factors that appear in both the numerator and the denominator in bold and red. Those are the ones that can be canceled. Leaving us with:

$$\frac{x + 4}{x - 4}\cdot\frac{x + 3}{x + 1}$$

Now we need to finish the calculation by multiplying the numerators and the denominators.

$$\frac{(x + 4)(x + 3)}{(x - 4)(x + 1)}=\frac{x^2 + 7x + 12}{x^2 - 3x - 4}$$

Because we canceled out 2x - 1 which equals 0 and x = 1/2, we also have to mention that our simplified version is undefined at that value.

Example 3

Multiply $\frac{x^2 - 4x + 3}{x^2 + 2x - 15}\cdot\frac{2x^2 + 13x + 15}{2x^2 + x - 3}$.

We'll start off again by factoring.

$$\frac{(x - 3)(x - 1)}{(x - 3)(x + 5)}\cdot\frac{(2x + 3)(x + 5)}{(2x + 3)(x - 1)}$$

In this case, all of the factors are in both the numerator and the denominator. Remember from Example 1 that whenever all the factors in a numerator or denominator get canceled, we replace them with a 1 so our expression becomes:

$$\frac{1}{1}=1$$

with the exception that x can't equal -5, 1, 3 or -3/2.

Dyanmic Tutorial - Multiplying Rational Expressions

Directions: This solution has 7 steps. To see a description of each step click on the boxes on the left side below. To see the calculations, click on the corresponding box on the right side. Try working out the solution yourself and use the descriptions if you need a hint and the calculations to check your solution.