Lecture : Multiplying Rational Expressions


 

In order to multiply two rational expressions, we first factor the numerator and the denominator of both expression.We then cancel any terms common to both a numerator and a denominator. We may cancel factors where one is on top of the other, we may also cancel like terms which lie diagonally from one another.

Here is an example:

 

(2x² + 5x + 3)
(2x² + 7x + 6)

x

(x² - 4)
(x3 + 1)


after factoring we will have:

 

(2x + 3) (x + 1)
(2x + 3)(x + 2)

x

    (x - 2)(x + 2) 
(x + 1) (x² + x + 1)

Since (2x + 3) occurs in both the numerator and the denominator, we can cancel it and we will have:
 

(x + 1)
(x + 2)

x

    (x - 2)(x + 2) 
(x + 1) (x² + x + 1)

(x + 1) also occurs in both the numerator and the denominator, so we cancel it also leaving us with:
 

    1 
(x + 2)

x

  (x - 2)(x + 2) 
(x² + x + 1)

(x-2) appears only in the numerator, so we may not cancel it, but (x + 2) appears in both the numerator and the denominator so we may cancel that:

  1 
  1 

x

     (x - 2) 
(x² + x + 1)

Now we multiply the numerators, and multiply the denominators to get our answer:

     (x - 2) 
(x² + x + 1)


 
 

Multiply each of the following.

1.

    (x² -16) 
(x²  + 7x + 12)

x

(x² - 4x - 21)
(x² - 4x)

    :solution

2.

(y² - 81)
 (y² - 16) 

x

(y² - y - 20)
(y² + 5y - 36)

    :solution

3.

(4x + 8)
(x³ -27)

x

(x²- 9)
(x²- 4)

    :solution

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#1 solution

    (x² -16) 
(x²  + 7x + 12)

x

(x² - 4x - 21)
(x² - 4x)

after factoring we will have:

 

(x + 4)(x - 4)
(x + 3)(x + 4)

x

(x - 7)(x + 3)
x(x - 4)

Now look at each factor in the numerators to see if it will cancel with any factors in the denominators.
Since (x + 4) occurs in both the numerator and the denominator, we can cancel it and we will have:
 

 (x - 4) 
(x + 3)

x

(x - 7)(x + 3)
x(x - 4)

(x - 4) also occurs in both the numerator and the denominator, so we cancel it also leaving us with:
 

    1 
(x + 3)

x

(x - 7)(x + 3)
x

(x - 7) appears only in the numerator, so we may not cancel it, but (x + 3) appears in both the numerator and the denominator so we may cancel that:

  1 
  1 

x

(x - 7)
x

Now we multiply the numerators, and multiply the denominators to get our answer:

(x - 7)
x

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#2 solution
 

(y² - 81)
 (y² - 16) 

x

(y² - y - 20)
(y² + 5y - 36)

after factoring we will have:

 

(y - 9)(y + 9)
(y - 4)(y + 4)

x

(y - 5)(y + 4)
(y + 9)(y - 4)

(y - 9) appears only in the numerator, so we may not cancel it, but (y + 9) appears in both the numerator and the denominator so we may cancel that:

    (y - 9) 
(y - 4)(y + 4)

x

(y - 5)(y + 4)
(y - 4)

(y - 5) appears only in the numerator, so we may not cancel it, but (y + 4) appears in both the numerator and the denominator so we may also cancel it:

(y - 9)
(y - 4)

x

(y - 5)
(y - 4)

(y - 9)(y - 5)
(y - 4)²

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#3 solution
 

(4x + 8)
(x³ -27)

x

(x²- 9)
(x²- 4)

after factoring we will have:

 

       4(x + 2)     
(x - 3)(x² + 3x + 9)

x

(x - 3)(x + 3)
(x + 2)(x - 2)

4 is not common to both a numerator and a denominator, so we may not cancel it. However, (x + 2) is, so we begin by cancelling it:
 
 

           4 
(x - 3)(x² + 3x + 9)

x

(x - 3)(x + 3)
(x - 2)

(x -3) also occurs in both the numerator and the denominator, so we cancel it also leaving us with:
 

        4 
(x² + 3x + 9)

x

(x + 3)
(x - 2)

Now, since we have no other common factors, we multiply the numerators, and multiply the denominators to get our answer:

        4 (x + 3) 
(x² + 3x + 9)(x - 2)

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