Precalc0006to0010-page23

Precalc0006to0010-page23 - ± 4 3 x ¸ ¹ º ¼ ½ 4 x 7...

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(Section 0.9: Simplifying Algebraic Expressions) 0.9.5 Method 1 (Factor the Numerator) We will factor out the GCF of the numerator, which is x 4 x + 7 () ± 2/3 . TIP 2 : It may be easier to factor out x in one step and then 4 x + 7 () ± 2/3 in another step. The GCF does not have to be factored out in one step. TIP 3 : It may be easier to substitute u = 4 x + 7 and obtain 2 xu 1/3 ± 4 3 x 2 u ± 2/3 u 2/3 . Make sure to substitute back later. 2 x 4 x + 7 () 1/3 ± 4 3 x 2 4 x + 7 () ± 2/3 4 x + 7 () 2/3 or 2 xu 1/3 ± 4 3 x 2 u ± 2/3 u 2/3 = x 4 x + 7 () ± 2/3 24 x + 7 () 1 3 ±± 2 3 ² ³ ´ µ · ± 4 3 x ¸ ¹ º » ¼ ½ 4 x + 7 () 2/3 or xu ± 2/3 2 u 1 3 ±± 2 3 ² ³ ´ µ · ± 4 3 x ¸ ¹ º º » ¼ ½ ½ u 2/3 = x 4 x + 7 () ± 2/3 24 x + 7 ()
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Unformatted text preview: ± 4 3 x ¸ ¹ º » ¼ ½ 4 x + 7 ( ) 2/3 or xu ± 2/3 2 u ± 4 3 x ¸ ¹ º » ¼ ½ u 2/3 We now divide 4 x + 7 ( ) ± 2/3 by 4 x + 7 ( ) 2/3 . If we let u = 4 x + 7 : 4 x + 7 ( ) ± 2/3 4 x + 7 ( ) 2/3 = u ± 2/3 u 2/3 = u ± 2 3 ± 2 3 = u ± 4 3 = 1 u 4/3 = 1 4 x + 7 ( ) 4/3 The division yields 4 x + 7 ( ) 4/3 in the denominator. = x 2 4 x + 7 ( ) ± 4 3 x ² ³ ´ µ ¶ · 4 x + 7 ( ) 4/3 = x 8 x + 14 ± 4 3 x ² ³ ´ µ ¶ · 4 x + 7 ( ) 4/3...
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