22
MULTIPLYING AND DIVIDING
ALGEBRAIC FRACTIONS
Section 2
TO MULTIPLY FRACTIONS, multiply the numerators and multiply the denominators, as in arithmetic.
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 |
| If a multiplication looks like this: 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! |
If any numerator has a divisor in common with any denominator,
they may be canceled.
| a b |
· | c d |
· | e a |
= | ce bd |
The a's cancel.
For if we took the trouble to multiply and write
| ace bda |
then it's obvious that we could divide both the numerator and denominator by a. It is more skillful, then, to reduce before multiplying.
Problem 3. Multiply. Reduce 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 reduce. Finally, multiply.
And remember: Only factors can be divided.
| 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 |
|||||
Section 2: Complex fractions -- Division
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