Question

How does factoring out variables with fractional exponents work? For example for (1/2)x^(1/2) - (1/2)x^(-1/2)

Answer 1

Doesn't matter. The thing you're factoring it from only has to be multiplied by what you're factoring out. It could be:

(1/2)x^(1/2)-(1/2)x^(apples) and you could still factor out the 1/2.

Answer 2

I just don't quite comprehend what I would get by factoring out a (1/2)x^(1/2) out of the (1/2)x^(-1/2) part.... then negative exponent confuses me

Answer 3

Oh I see, I misread your question earlier.

Answer 4

So whenever you factor something out, think of it kind of like this:

\[\frac{ x^{1/2} }{ 2 }-\frac{1 }{ 2x^{1/2} }\]

Now multiply by a fancy form of 1, since anything divided by itself is just 1.

\[\frac{ \frac{ x^{1/2} }{ 2 } }{\frac{ x^{1/2} }{ 2 } }*(\frac{ x^{1/2} }{ 2 }-\frac{1 }{ 2x^{1/2} })\]

See, now the thing you were factoring out is on top, and distribute the bottom part. This is much easier than trying to pull something out of it.

\[\frac{ x^{1/2} }{ 2 } [\frac{ 2 }{ x^{1/2} }*(\frac{ x^{1/2} }{ 2 }-\frac{1 }{ 2x^{1/2} })]\]
Are you with me so far? I just flipped the fraction upside down and I'm ready to distribute while the thing I factored is still on the left.

\[\frac{ x^{1/2} }{ 2 } (1-\frac{2 }{ 2*x^{1/2}*x^{1/2} })\]\[\frac{ x^{1/2} }{ 2 } (1-\frac{1 }{ (x^{1/2}) ^2 })\]\[\frac{ x^{1/2} }{ 2 } (1-\frac{1 }{ x })\]

Hopefully that gives you an easier way to think about factoring.

Answer 5

What? That's completely different.

Answer 6

whoops, wrong copy paste :P
(1/2)x^(1/2) - (1/2)x^(-1/2) get
((1/2)x^(-1/2))(x-1) ?

Answer 7

Well, you tell me. Distribute the left to the right and see if you get it to check if that's the correct factorization.