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Binomial Expansion

I have the fraction 2x−15+2x \frac{2x-1}{\sqrt {5+2x}} .

Now, I have the solution to the binomial expansion of 15+2x \frac{1}{\sqrt{5+2x}} , and I need to use it to find the solution to the formerly mentioned fraction. I believe I could do this if I understood how to get this fraction in a form similar to the first, or in the form (a+bx)n(a+bx)^n . The problem lies in the fact that I don't know how to do this when both the numerator and denominator of the fraction are algebraic. Could explain how I would convert this?
Original post by Illidan2
I have the fraction 2x−15+2x \frac{2x-1}{\sqrt {5+2x}} .

Now, I have the solution to the binomial expansion of 15+2x \frac{1}{\sqrt{5+2x}} , and I need to use it to find the solution to the formerly mentioned fraction. I believe I could do this if I understood how to get this fraction in a form similar to the first, or in the form (a+bx)n(a+bx)^n . The problem lies in the fact that I don't know how to do this when both the numerator and denominator of the fraction are algebraic. Could explain how I would convert this?


The expansion you're looking for is (2x-1) times the one you've already found.

So, if you have a+bx+cx2+dx3...a +bx+cx^2 +dx^3..., multiplying by 2x-1 we have (2x−1)(a+bx+cx2+dx3...)(2x-1)(a +bx+cx^2 +dx^3...) and simply expand the brackets.
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Illidan2
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Original post by ghostwalker
The expansion you're looking for is (2x-1) times the one you've already found.

So, if you have a+bx+cx2+dx3...a +bx+cx^2 +dx^3..., multiplying by 2x-1 we have (2x−1)(a+bx+cx2+dx3...)(2x-1)(a +bx+cx^2 +dx^3...) and simply expand the brackets.


Of course! Now that you've said it, it's become so clear. Can't believe I missed something so obvious. Thank you! :smile:

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