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Multiplying Rational Expressions | Step by Step

Rational expressions are multiplied in the same way as you would with regular fractions. As you may have learned, we multiply simple fractions using the steps below...

simple illustration showing how to mutliply fractions: (9/5)*(5/4) = 9/4

 

Steps in Multiplying Rational Expressions

  • Multiply the numerators together
  • Multiply the denominators together
  • Simplify the "new" fraction by cancelling common factors when possible

Let us proceed by going over five (5) worked examples in this lesson.


Direction: Simplify the given rational expressions by multiplication.

1) example 1: (2x+1)/(x^2-1) * (x+1)/(2x^2+x) See solution
2) example 2: (x^2+7x+10)/(x^2-9x+14) * (x^2-6x-7)/(x^2+6x+5) See solution
3) example 3: (8x^2-6x)/(4-x) * (x^2-16)/(4x^2+x-3) * (-5x-5)/(2x+8) See solution
4) example 4: (x^5)/(4x^3-3x^2) * (4x^2+5x-6)/(x^2-2x-35) * (x^2+10x+25)/(x^2+7x+10) See solution
5) example 5: (-3x^2+27)/(x^3-1)*(7x^3+7x^2+7x)/(x^2-3x)*(x-1)/21 See solution

 


Example 1: Multiply the rational expressions   (2x+1)/(x^2-1) * (x+1)/(2x^2+x).

I see that both denominators are factorable. The first denominator is a case of difference of two squares. The second denominator is easy because I can pull out a factor of x.

This is how it looks.

x^2-1 = (x+1)(x-1), and 2x^2+x = x(2x+1)

 

=(2x+1)/[(x+1)(x-1)] * (x+1)/[x(2x+1)]   Factor the denominators
=[(2x+1)(x+1)]/[(x+1)(x-1)(x)(2x+1)]  

Now, I can multiply across the numerators and across the denominators by placing them side by side.

At this point, I compare the top and bottom factors and decide which ones can be crossed-out.

cancel common factors: (2x+1) and (x+1)  

By color coding the common factors, it is clear which ones to eliminate.

The factors (2x+1) and (x+1) should both be cancelled out as shown.

(2x+1)/(x^2-1) * (x+1)/(2x^2+x)=1/[x(x-1)]  

What remains on top is just the number 1.

And so we have this as our final answer.

Try not to distribute it back and keep it in factored form. However, if your teacher wants the final answer to be distributed, then do so. Either case should be correct. It's just a matter of preference.

 


Example 2: Multiply the rational expressions   (x^2+7x+10)/(x^2-9x+14) * (x^2-6x-7)/(x^2+6x+5).

We need to factor out all the trinomials. The good news is that this type of trinomial, where the coefficient of the squared term is 1, is very easy to handle. Starting with the first numerator, find two numbers where its product gives the last term (10) and its sum gives the middle coefficient (7). I'm thinking of +5 and +2. They should work, agree? For the second numerator, the two numbers must be −7 and +1 since their product is the last term (−7) while its sum is the middle coefficient (−6).

To factor the first denominator, find two numbers with a product of the last term (14) and a sum of the middle coefficient (-9). By trial and error, the numbers are −2 and −7. However, you should always verify it. Now for the second denominator, think of two numbers when multiplied gives the last term (5) and when added gives (6). Obviously they are +5 and +1.

The correct factors for each trinomial should appear like this.

factor out all the expressions in both numerator and denominator: x^2+7x+10=(x+5)(x+2), x^2-6x-7=(x-7)(x+1), x^2-9x+14=(x-2)(x-7), and x^2+6x+5=(x+5)(x+1)

 

=[(x+5)(x+2)]/[(x-2)(x-7)] * [(x-7)(x+1)/(x+5)(x+1)]   Factor the numerators and denominators completely.
=[(x+5)(x+2)(x-7)(x+1)]/[(x-2)(x-7)(x+5)(x+1)]  

Multiply them together - numerator against numerator and denominator against denominator.

In other words, place side by side in a single fractional symbol.

cancel out the common factors: (x+5), (x-7) and (x+1)   The color schemes should aid in identifying common factors that we can get rid of.
(x^2+7x+10)/(x^2-9x+14) * (x^2-6x-7)/(x^2+6x+5) = (x+2)/(x-2)  

Yep! We cleaned it out beautifully. This is our final answer.

 


Example 3: Multiply the rational expressions  (8x^2-6x)/(4-x) * (x^2-16)/(4x^2+x-3) * (-5x-5)/(2x+8).

In this problem, I have six terms that need factoring. However, most of them are easy to handle and I will provide suggestions on how to factor each. I will also show a quick side calculation how to factor 4x2+x-3 because it can be challenging to some.

factor out the difference of two equares   factor out the trinomial: 4x^2+x-3 = (4x-3)(x+1)

 

all expressions have been factored out completely   Factoring out all the terms.
=[2x(4x-3)(x+4)(x-4)(-5)(x+1)]/[(-1)(x-4)(x+1)(4x-3)(2)(x+4)]  

Multiply all of them at once by placing them side by side. All numerators stay on top and denominators at the bottom.

cancel out the common factors (4x-3) and (x-4)  

I decide to cancel common factors one or two at a time so that I can keep track of them accordingly. Otherwise, I may commit "careless" errors. 

I will first get rid of the terms 4x−3 and x−4.

=[2x(x+4)(-5)(x+1)]/[(-1)(x+1)(2)(x+4)]   Simplify
cancel out common factors again: (x+4) and (x+1)   Next, I will eliminate the factors x+4 and x+1.
=[2x(-5)]/(-1)(2)   Simplify. Now I am thinking to cancel the 2's.
one more cancellation of the factor 2x   Let's do that.
=x(-5)/ (-1)  

At this point, there's really nothing to cancel. However there's something I can simplify by division.

Notice that (−5) ÷ (−1) = 5.

(8x^2-6x)/(4-x) * (x^2-16)/(4x^2+x-3) * (-5x-5)/(2x+8)=5x  

There  you go! This is our final answer.

 


Example 4: Multiply the rational expressions  (x^5)/(4x^3-3x^2) * (4x^2+5x-6)/(x^2-2x-35) * (x^2+10x+25)/(x^2+7x+10).

 

As you can see, there are so many things going on in this problem. However, don't be intimidated by how it looks. Start by factoring each term completely. The problem will become easier as you go along.

factor out all the trinomials: 4x^2+5x-6 =(x+2)(4x-3), x^2+10x+25=(x+5)(x+5), x^2+7x+10=(x+5)(x+2), x^2-2x-35=(x-7)(x+5)

In fact, once we have factored the terms correctly, the rest of the steps becomes manageable.

=multiply together all the numerators. do the same with all the denminators.   Factor out each term completely.
=[x^5(x+2)(4x-3)(x+5)(x+5)/[(x^2(4x-3)(x-7)(x+5)(x+5)(x+2)]  

Multiply by placing them in a single fractional symbol. All numerators are written side by side on top while the denominators at the bottom.

cancel out the common factors: (x+5)   I will first cancel all the (x+5) terms.
=[x^5(x+2)(4x-3)]/[x^2(4x-3)(x-7)(x+2)]   Simplify
cancel out again the common factors (x+2) and (4x-3)   Next, cross out the (x+2) and (4x−3) terms.
=x^5/[x^2(x-7)]   At this point, I can also simplify the monomials with variable x.
x^5/x^2= x^3  

There are five x's on top and two x's at the bottom.

Cancelling the x's with one for one correspondence should leave us three x's in the numerator.

(x^5)/(4x^3-3x^2) * (4x^2+5x-6)/(x^2-2x-35) * (x^2+10x+25)/(x^2+7x+10)=x^3/(x-7)  

This is the final answer.

Caution: Don't do this! don't cancel the x here: x^5/ (x-7)

Note that the x in the denominator is not by itself. It is part of the entire term x−7. This is a common error of many students. Don't fall into this common mistake.


Example 5: Multiply the rational expressions   (-3x^2+27)/(x^3-1)*(7x^3+7x^2+7x)/(x^2-3x)*(x-1)/21.

I am sure that by now, you are getting better on how to factor. The only thing I need to point out is the denominator of the first rational expression, x3−1. This is a special case called difference of two cubes.

formula to factor sum of two cubes: a^3+b^3 = (a+b)(a^2-ab+b^2), and factoring out difference of two cubes: a^3-b^3 = (a-b)(a^2+ab+b^2)

In this problem, I will use Case 2 because of the "minus"; hence the word difference. Case 1 is known as the sum of two cubes because of the "plus" symbol.

These are the factors...

factor each expression: -3x^2+27 =-2(x+3)(x-3), 7x^3+7x^2+7x=7x(x^2+x+1), x^3-1 =(x-1)(x^2+x+1) and x^2-3x =x(x-3)

 

everything factored out before multiplying the rational experessions   Factorize all the terms as much as possible.
=[-3*(x+3)(x-3)*7x(x^2+x+1)(x-1)/[(x-1)(x^2+x+1)(x)(x-3)(21)]  

Multiply the numerators together and do the same with the denominators.

cancel out the common trinomial factor x^2+x+1   I will first cross out the term x2+x+1.
=[-3*(x+3)(x-3)(7x)(x-1)]/[(x-1)(x)(x-3)(21)]   Next, I will eliminate the terms (x−1) and (x−3).
cancel common factors: (x-3) and (x-1)   Here it goes. I hope the color coding help you keep track which terms are being cancelled out.
=(-3)(x+3)(7x)/(x*21)   I see a single x term on both the top and bottom.
=-3*(x+3)(7)/x*21   Cross out that x as well.
cancel out x   At this point, I will multiply the constants on the numerator.
=-21*(x+3)/21   Since (−3) (7) = −21.
cancel out -21   Cancel the 21's but leaving −1 on top.
=-1*(x+3)   I can keep this as the final answer. However, it will look better if I distribute −1 into (x+3).
(-3x^2+27)/(x^3-1)*(7x^3+7x^2+7x)/(x^2-3x)*(x-1)/21=-x-3  

We got it again!

 


 

Practice Problems with Answers
Worksheet 1 Worksheet 2

 

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