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A L G E B R A

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# MULTIPLYING AND DIVIDINGALGEBRAIC FRACTIONS

The rule

Section 2

Complex fractions -- Division

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) 2x · 5x = 10x2 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 − 3x + 1 · x − 2x + 1 = x2 − 5x + 6x2 + 2x + 1
 If a multiplication looks like this:  a· bc or bc ·  a,  multiply only

the numerator.

 a· bc = ab c

Problem 2.   Multiply.

 a) x · 2x 3 = 2x2 3 b) 3x2 4 ·  7x3 = 21x5  4
 c)   (x + 3)· x − 3x + 6 = x2 − 9 x + 6
 d) x2 − 2x + 5 6x2 − 4x + 1 ·  2x3 = 6x2 − 4x + 1 No canceling!

Reducing

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

 ab · cd · ea = cebd

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) abcd · ed fg · hcfake = bhgk
 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· bcq = ab c f)   10· x + 2   2 = 5(x + 2) =  5x + 10
 g)  3x· 5x 6 = 5x2 2
 h) −a b · 1a = − 1b
 The a's cancel as −1, which on multiplication with 1 makes the fraction itself negative (Lesson 4).
 Example 1.   Multiply x2 − 4x − 5x2 − x − 6 · x2 − 5x + 6x2 − 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 − 5x2 − x − 6 · x2 − 5x + 6x2 − 6x + 5 = (x + 1)(x − 5)(x + 2)(x − 3) · (x − 3)(x − 2)(x − 1)(x − 5) = x + 1x + 2 · x − 2x − 1 = x2 − x − 2x2 + 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 + 1x2 − x − 12 · x2 + x − 6x2 − 6x + 5 = __(x − 1)2__(x − 4)(x + 3) · (x + 3)(x − 2)(x − 1)(x − 5) = x − 1x − 4 · x − 2x − 5 = x2 − 3x + 2x2 − 9x + 20
 c) x2 + 3x − 10x2 + 4x − 12 · x2 + 5x − 6x2 + 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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