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22

MULTIPLYING AND DIVIDING
ALGEBRAIC FRACTIONS

The rule



TO MULTIPLY FRACTIONS, multiply the numerators and multiply the denominators.

Problem 1.   Multiply.

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Do the problem yourself first!

  a)    2
x
·   5
x
  =   10
x²
  b)    3ab
 4c
·   4a²b
 5d
  =   3a³b²
 5cd
   The 4's cancel.
  c)      3x  
x + 1
·    6x² 
x − 1
  =    18x³ 
x² − 1
   The Difference of Two Squares
  d)    x − 3
x + 1
·   x − 2
x + 1
  =   x² − 5x + 6
x² + 2x + 1
In a multiplication of this form  a·   b
c
  or    b
c
·  a,   multiply only

the numerator.


a·   b
c
 =   ab
 c

Problem 2.   Multiply.

  a)    x
·   2x
 3
  =   2x²
 3
  b)    3x²
 4
·  7x3   =  
  c)   (x + 3)·   x − 3
x + 6
  =   x² − 9
 x + 6
  d)     x² − 2x + 5 
6x² − 4x + 1
·  2x3   =  
  6x² − 4x + 1
   No canceling!

Canceling

If any numerator has a divisor in common with any denominator,
it may be canceled.

a
b
·    c
d
·   e
a
  =    ce
bd

The a's cancel.

Problem 3.   Multiply.  Cancel first.

  a)   ab
cd
·    ed
 fg
·    hcf
ake
  =   bh
gk
  b)    (x − 2)(x + 2)
        8x
·        __2x__     
(x + 2)(x − 1)
  =      x − 2 
4(x − 1)
  c)         __x³__     
(x + 2)(x + 3)
·   x + 3
  x7
  =     __1_   
(x + 2)x4
  d)    x(x + 1)
     6
·       2    
x² − 1
  =   x(x + 1)
     6
·        __2__     
(x + 1)(x − 1)
  =      _x_   
3(x − 1)
  e)   aq·   b
cq
  =   ab
 c
  f)   10·  x + 2
   2
  =   5(x + 2)   =  5x + 10
  g)  3x·  5x
 6
  =   5x²
 2
  h)    a
 b
·   1
a
  =   1
b
  The a's cancel as −1, which on multiplication with 1 makes the fraction itself negative (Lesson 4).
  Example 1.   Multiply     x² − 4x − 5
x² − x − 6
·   x² − 5x + 6
x² − 6x + 5

Solution.   Although the problem says "Multiply," that is the last thing to do in algebra!  First factor.  Then cancel.  Finally, multiply.

And remember:  Only factors cancel.

x² − 4x − 5
x² − x − 6
·   x² − 5x + 6
x² − 6x + 5
  =   (x + 1)(x − 5)
(x + 2)(x − 3)
·   (x − 3)(x − 2)
(x − 1)(x − 5)
 
    =   x + 1
x + 2
·   x − 2
x − 1
 
 
    =   x² − x − 2
x² + x − 2

Problem 4.   Multiply.

  a)       __x²__   
x² + x − 12
·   x² − 9
  2x6
  =        __x²__     
(x + 4)(x − 3)
·   (x − 3)(x + 3)
       2x6
 
    =      1   
x + 4
·   x + 3
  2x4
 
 
    =   _ x + 3 _
2x5 + 8x4
  b)    x² − 2x + 1
x² − x − 12
·   x² + x − 6
x² − 6x + 5
  =   __(x − 1)²__
(x − 4)(x + 3)
·   (x + 3)(x − 2)
(x − 1)(x − 5)
 
    =   x − 1
x − 4
·   x − 2
x − 5
 
 
    =   x² − 3x + 2
x² − 9x + 20
  c)    x² + 3x − 10
x² + 4x − 12
·   x² + 5x − 6
x² + 4x − 5
  =   (x + 5)(x − 2)
(x + 6)(x − 2)
·   (x − 1)(x + 6)
(x − 1)(x + 5)
 
    =   1  
  d)     _x³_ 
x² − 1
·   x² + x − 2
      x4
·    __x²__ 
x² + 4x + 4
  =      ___x³___   
(x + 1)(x − 1)
·   (x − 1)(x + 2)
       x4
·    __x²__
(x + 2)²
  =  
 x + 1
·      1   
x + 2
 
    =   _    _x_   _
x² + 3x + 2

Complex fractions:  Division

  A complex fraction looks like this:   

The numerator and/or the denominator are themselves fractions.

To deal with a complex fraction, we immediately apply the definition of division (Lesson 6):

a
b
  =  a·  1
b

Any fraction is equal to the numerator times the reciprocal
of the denominator.

Therefore,

  =   p
q
·   n
m
  Example 2.   Simplify   
  Solution.      =   x² − 25
   x8
·      x³ 
x − 5
    =   (x + 5)(x − 5)
        x8
·      x³ 
x − 5
 
    =   x + 5
   x5

Division -- which effectively this is -- becomes multiplication by the reciprocal.

on canceling the x + 2's.

Problem 5.   Simplify.

  a)       =    6 
x5
·   x²
8
  =     3  
4x3
  b)       =      4   
x − 1
·      1   
x − 1
  =         _4_      
x² − 2x + 1
  c)       =   (x + 2) ·   x − 2
x + 2
  =   x − 2
  d)       =   x + 2
x + 1
·   x − 2
x − 1
  =   x² − 4
x² − 1
  e)       =       −h    
x(x + h)
·   1
h
  =   −       1     
x(x + h)

The h's cancel.  And according to the Rule of Signs, the product is negative.  (It's all right to leave the product in its factored form.)

  f)       =   (x + 2)(x − 2)
       3x²
·          _x_       
(x + 2)(x + 3)
  =   x − 2
  3x
·      1   
x + 3
  =      x − 2   
3x(x + 3)

  Example 4.   Simplify   

Solution.    1-over any number is its reciprocal.  Therefore,

= 4
3

Problem 6.   Simplify the following.

  a)      =   x + 1
   x
  b)      =   x − 1

Next Lesson:  Adding algebraic fractions


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