 |  |
22
MULTIPLYING AND DIVIDING
ALGEBRAIC FRACTIONS
TO MULTIPLY FRACTIONS, multiply the numerators and multiply the denominators, as in arithmetic.
Problem 1. Multiply.
To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
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 | = | 21x5 4 |
| 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 5):
| a b |
= a· | 1 b |
Any fraction is equal to the numerator times the reciprocal
of the denominator.
Therefore,
|
= | p q |
· | n m |
Problem 5. State in words how to deal with a complex fraction.
Rewrite it as the numerator times the reciprocal
of the denominator.
| 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 6. 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 7. Simplify the following.
| a) |  |
= | x + 1 x |
b) |  |
= | x − 1 |
Next Lesson: Adding algebraic fractions
Please make a donation to keep TheMathPage online.
Even $1 will help.
Copyright © 2012 Lawrence Spector
Questions or comments?
E-mail: themathpage@nyc.rr.com