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Binomial expansion in the form of (a + x)^n

I just would like to check my work since I can not find any tool online to do so.
I must expand 16/(2+x)^2 in ascending power of x up to x^2 and simplify.
I found 4 - 4x + 3x^2, is that correct?
Reply 1
Avatar for mqb2766
mqb2766
21
Yes, and
https://www.wolframalpha.com/input/?i=Expand+16%2F%282%2bx%29%5E2+
(edited 3 years ago)
Original post by Mlopez14
I just would like to check my work since I can not find any tool online to do so.
I must expand 16/(2+x)^2 in ascending power of x up to x^2 and simplify.
I found 4 - 4x + 3x^2, is that correct?


Yes.

Wolfram is useful for checking - see here
Reply 3
Avatar for dumb2020
dumb2020
6
Original post by ghostwalker
Yes.

Wolfram is useful for checking - see here


I don't understand, since the quadratic is the denominator how can we get those terms
Reply 4
Avatar for Mlopez14
Mlopez14
OP
13
Original post by mqb2766

Thank you and thank you really much
Original post by dumb2020
I don't understand, since the quadratic is the denominator how can we get those terms


Have you covered the binomial expansion with negative indices?

Here you're expanding 16(2+x)−216(2+x)^{-2}
(edited 3 years ago)
Reply 6
Avatar for dumb2020
dumb2020
6
Original post by ghostwalker
Have you covered the binomial expansion with negative indices?

Here you're expanding 16(2+x)−216(2+x)^{-2}

Ohh right. I noticed their answer was the same as the taylor series from the link but I wasn't convinced I had to do it that way when the title said binomial expansion. Thanks!
Reply 7
Avatar for mqb2766
mqb2766
21
Original post by dumb2020
Ohh right. I noticed their answer was the same as the taylor series from the link but I wasn't convinced I had to do it that way when the title said binomial expansion. Thanks!

In a sense you could just solve
16 = (4 + 4x + x^2)(a + bx + cx^2 ....)
To get
a = 4
b = -4
c = 3
...
Without any Taylor/binomial "baggage".
Reply 8
Avatar for dumb2020
dumb2020
6
Original post by mqb2766
In a sense you could just solve
16 = (4 + 4x + x^2)(a + bx + cx^2 ....)
To get
a = 4
b = -4
c = 3
...
Without any Taylor/binomial "baggage".

Yeah that's a nice way too. Thank you!

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