besnuffelfo

besnuffelfo

Answered

2022-09-24

How to simplify this fraction with different powers?
I happen to be stuck trying to simplify this:
[ ( 3 x + 2 ) ( x + 1 ) 3 2 ( 3 2 x 2 + 2 x ) ( 3 2 ) ( x + 1 ) 1 2 ( x + 1 ) 3 ]

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Answer & Explanation

kregde84

kregde84

Expert

2022-09-25Added 10 answers

We have
( 3 x + 2 ) ( x + 1 ) 3 2 ( 3 2 x 2 + 2 x ) ( 3 2 ) ( x + 1 ) 1 2 ( x + 1 ) 3
divide numerator and denominator by ( x + 1 ) 1 2 giving
( 3 x + 2 ) ( x + 1 ) ( 3 2 x 2 + 2 x ) ( 3 2 ) ( x + 1 ) 5 2
multiply the ( 3 x + 2 ) ( x + 1 ) to get 3 x 2 + 5 x + 2 and the ( 3 2 x 2 + 2 x ) ( 3 2 ) to get 9 4 x 2 + 3 x giving:
3 x 2 + 5 x + 2 9 4 x 2 3 x ( x + 1 ) 5 2
then simplify the numerator:
3 x 2 + 5 x + 2 9 4 x 2 3 x = 12 4 x 2 9 4 x 2 + 5 x 3 x + 2 = 3 4 x 2 + 2 x + 2
to get
3 4 x 2 + 2 x + 2 ( x + 1 ) 5 2

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demitereur

demitereur

Expert

2022-09-26Added 2 answers

( 3 x + 2 ) ( x + 1 ) 3 2 ( 3 2 x 2 + 2 x ) ( 3 2 ) ( x + 1 ) 1 2 ( x + 1 ) 3 =
take the common factor ( x + 1 ) 1 2
= [ ( 3 x + 2 ) ( x + 1 ) ( 3 2 x 2 + 2 x ) ( 3 2 ) ] ( x + 1 ) 1 2 ( x + 1 ) 3 =
= ( 3 x + 2 ) ( x + 1 ) ( 3 2 x 2 + 2 x ) ( 3 2 ) ( x + 1 ) 3 ( x + 1 ) 1 2 = ( 3 / 4 ) x 2 + 2 x + 2 ( x + 1 ) 5 2

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