# Factoring algebraic expressions contaning fractions

• B

## Main Question or Discussion Point

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.

Mark44
Mentor
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.

• leighflix
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

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.

SteamKing
Staff Emeritus
Homework Helper
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. • leighflix
SteamKing
Staff Emeritus
Homework Helper
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: $a ⋅ b^{-n} = \frac{a}{b^n}$

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.

SteamKing
Staff Emeritus
• Thanks both of you! 