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:

 

after factoring we will have:

 

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


 

 

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

 

 

(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:

 

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


 
 

Multiply each of the following.

1.

 

:solution

 

2.

 

:solution

 

3.

 

:solution

 

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#1 solution

 

after factoring we will have:

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

(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:

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

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#2 solution
 

after factoring we will have:

(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 - 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 - 5)
(y - 4)2

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#3 solution
 

after factoring we will have:

 

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:
 
 

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

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

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Lecture : Dividing Rational Expressions


 

Recall that rational expressions are essentially fractions, the difference being that they have polynomials as their numerators and denominators rather than the integers we are more familiar with. When we divide with those fractions, we invert the divisor and multiply (or as it is commonly known "flip the second fraction" and multiply, or multiply by the reciprocal of the denominator)

Ex)

 3 
10

 9 
25

becomes

 3 
10

x

25
9

Which after cancelling becomes

1
2

x

5
3

And The answer is

5
6

With rational expressions we follow the same steps, we invert the divisor and multiply.

Here is an example:

 

First we will invert the second fraction and change the operation to multiplication

after factoring we will have:

Notice x , (x + 1) and (x + 2) occur in both the numerator and the denominator:
 

   5x(x + 2)
x(x - 2)(x + 2)

´

5(x + 1)(x + 1)
(x - 1)(x + 1)

So we can cancel them and get:

   5 
(x - 2)

´

5(x + 1)
(x - 1)

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

   25(x + 1) 
(x - 2)(x - 1)


 

Divide each of the following.

1. 

 

 (5x2 + 7x + 2) 
(12x2 -19x - 21)

(8x2+ 13x + 5)
(6x2 + x - 35)

    :solution

2. 

 

(x2 + 10x)
(x2 - 1)

(x3 + 1000)
(5x2 + 6x + 1)

    :solution

3. 

 

(20x2 + 64x - 21)
(6x - 21)

(4x2 -49)

    :solution

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#1 solution

 (5x2 + 7x + 2) 
(12x2 - 19x - 21)

(8x2 + 13x + 5)
(6x2 + x - 35)

First we will invert the second fraction and change the operation to multiplication
 

 (5x2+ 7x + 2) 
(12x2 - 19x - 21)

x

(6x2 + x - 35)
(8x2 + 13x + 5)

after factoring we will have:

 

(5x + 2)(x + 1)
(4x + 3)(3x - 7)

x

(3x - 7)(2x + 5)
(x + 1)(8x + 5)

Notice that (x + 1) and (3x - 7) occur in both the numerator and the denominator:

(5x + 2)(x + 1)
(4x + 3)(3x - 7)

x

(3x - 7)(2x + 5)
(x + 1)(8x + 5)

So we can cancel them and have:

(5x + 2)
(4x + 3)

x

(2x + 5)
(8x + 5)

There are no more common factors so we multiply the numerators, and multiply the denominators to get our answer:

(5x + 2)(2x + 5)
(4x + 3)(8x + 5)


 

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#2 solution
 

(x2 + 10x)
(x2 - 1)

(x3 + 1000)
(5x2 + 6x + 1)

First we will invert the second fraction and change the operation to multiplication
 

(x2 + 10x)
(x2 - 1)

´

(5x2 + 6x + 1)
(x3 + 1000)

after factoring we will have:

 

x (x + 10 )
(x + 1)(x - 1)

´

    (5x + 1)(x + 1) 
(x + 10)(x2 - 10x + 100)

Since (x + 10) and (x + 1) occurs in both the numerator and the denominator, we can cancel them:

x (x + 10 )
(x + 1)(x - 1)

´

    (5x + 1)(x + 1) 
(x + 10 )(x2 - 10x + 100)

   x 
(x - 1)

´

    (5x + 1) 
(x2 - 10x + 100)

There are no more common factors so we multiply the numerators, and multiply the denominators to get our answer:

      x(5x + 1) 
(x - 1)(x2 - 10x + 100)

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#3 solution

3. 

 

(202 + 64x - 21)
(6x - 21)

(4x2 - 49)

   

Note that (4x² - 49) does not have a denominator, we may put it over 1, so that it is in fraction form.

Now we will invert the second fraction and change the operation to multiplication
 

after factoring we will have:

 

(2x + 7)(10x – 3)
3(2x - 7)         1 
(2x + 7)(2x – 7)

Since (2x + 7) occurs in both the numerator and the denominator, we can cancel it and have:

(10x – 3)
3(2x – 7)

´

      1 
(2x – 7)

There are no more common factors so we multiply the numerators, and multiply the denominators to get our answer:

(10x – 3)
3(2x – 7)2

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