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

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Mlopez14
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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?
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mqb2766
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Yes, and
https://www.wolframalpha.com/input/?i=Expand+16%2F%282%2bx%29%5E2+
Last edited by mqb2766; 1 month ago
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ghostwalker
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(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
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dumb2020
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(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
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Mlopez14
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Thank you and thank you really much
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ghostwalker
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(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)^{-2}
Last edited by ghostwalker; 1 month ago
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dumb2020
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(Original post by ghostwalker)
Have you covered the binomial expansion with negative indices?

Here you're expanding 16(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!
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mqb2766
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(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".
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dumb2020
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(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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