Factoring algebraic expressions contaning fractions

  • #1
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http://imgur.com/RNsBBoO (image)

Can someone elaborate as to how he factored this? The textbook provided nothing else about factoring algebraic expressions with fractions than this image.It avoided fractions in factoring like a plague I guess.

I understand the 3rd step to put the (x + 1)^(3/4) on the bottom since the exponent was negative. However I have absolutely no idea how the former 2 steps were possible.
 

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  • #2
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http://imgur.com/RNsBBoO (image)

Can someone elaborate as to how he factored this? The textbook provided nothing else about factoring algebraic expressions with fractions than this image.It avoided fractions in factoring like a plague I guess.

I understand the 3rd step to put the (x + 1)^(3/4) on the bottom since the exponent was negative. However I have absolutely no idea how the former 2 steps were possible.
Do you agree that ##(x + 1)^{1/4}## in the first step is the same as (equal to) ##(x + 1)^{-3/4}(x + 1)## in the second step?
The purpose of doing this was to get a common factor of ##(x + 1)^{-3/4}##, and then using the distributive law.
 
  • #3
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Do you agree that ##(x + 1)^{1/4}## in the first step is the same as (equal to) ##(x + 1)^{-3/4}(x + 1)## in the second step?
The purpose of doing this was to get a common factor of ##(x + 1)^{-3/4}##, and then using the distributive law.
I have no idea how (x + 1)^(1/4) = (x + 1)^(-3/4) * (x + 1)

EDIT: Ok, I get how it is equilavent
 
  • #4
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In step 1 & 2, did he group factor?

(2x + 1)(x + 1)^(-3/4) = (2x+1) / (x + 1)^(3/4)

EDIT: OK no, he didn't group factor, he basically just added x and (x+1).
I understand now, thanks.
 
  • #5
SteamKing
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I have no idea how (x + 1)^(1/4) = (x + 1)^(-3/4) * (x + 1)

EDIT: Ok, I get how it is equilavent
It's also equivalent. :wink:
 
  • #6
SteamKing
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In step 1 & 2, did he group factor?

(2x + 1)(x + 1)^(-3/4) = (2x+1) / (x + 1)^(3/4)
Naw. With a negative exponent, you can just re-write it so that term appears in the denominator.

You know: ## a ⋅ b^{-n} = \frac{a}{b^n}##
 
  • #7
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Naw. With a negative exponent, you can just re-write it so that term appears in the denominator.

You know: ## a ⋅ b^{-n} = \frac{a}{b^n}##
Yea, that was what I was thinking. Simply just add similar to adding fractions, since they have the same denominator, just add both numerators.
 
  • #8
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Yea, that was what I was thinking. Simply just add similar to adding fractions, since they have the same denominator, just add both numerators.
You got it.
 
  • #9
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Thanks both of you! :smile:
 

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