Factoring algebraic expressions contaning fractions

[leighflix]

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]

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

 

[leighflix]

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]

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.

 

[SteamKing]

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: $a ⋅ b^{-n} = \frac{a}{b^n}$

 

[leighflix]

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]

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.

 

[leighflix]

Thanks both of you!