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Coefficient of x^3

by Stacyg
Tags: coefficient
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Stacyg
#1
May15-08, 06:06 PM
P: 25
Find the coefficient of x^3 in the expansion of (2x^2-3/x)^3


I know how to do simple coefficients using pascalles triangle but I really don't know how to do this.
Any help would be much appreciated.
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Tedjn
#2
May15-08, 06:08 PM
P: 738
Write it as

[tex](2x^2 - 3x^{-1})^3[/tex]

From Pascal's triangle, you know how to expand

[tex](a+b)^n[/tex]

What can you replace with a and what can you replace with b?
rock.freak667
#3
May15-08, 06:09 PM
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P: 6,202
[tex](2x^2-\frac{3}{x})^3[/tex]

[tex] (\frac{1}{x}(2x^3-3))^3[/tex]

How about now?

Tedjn
#4
May15-08, 06:10 PM
P: 738
Coefficient of x^3

Oh, that's a nice way of doing it :)
rock.freak667
#5
May15-08, 07:36 PM
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Quote Quote by Tedjn View Post
Oh, that's a nice way of doing it :)
Usually (well for me), a binomial expansion is usually done with a variable and a constant.

as for [itex](a+b)^n[/itex] is valid for [itex]|\frac{b}{a}|<1[/itex] But if a and b are variables, you'll have to do some fancy algebra to get the range for which it is valid.
Dick
#6
May15-08, 08:23 PM
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Quote Quote by rock.freak667 View Post
Usually (well for me), a binomial expansion is usually done with a variable and a constant.

as for [itex](a+b)^n[/itex] is valid for [itex]|\frac{b}{a}|<1[/itex] But if a and b are variables, you'll have to do some fancy algebra to get the range for which it is valid.
Why is it only valid in some range??? I also don't see why you need to factor the original. (a+b)^3=a^3+3*a^2*b+3*a*b^2+b^3. Just put a=2x^2 and b=(-3/x), figure out which term is the x^3 term and evaluate it.
rock.freak667
#7
May15-08, 08:53 PM
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That's what I was taught.."validity of a binomial"
Dick
#8
May15-08, 10:30 PM
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Quote Quote by rock.freak667 View Post
That's what I was taught.."validity of a binomial"
Got a reference? If you are thinking of the convergence of the infinite series for negative exponents, that is something to think about. But this is a positive exponent, the series is finite. There are no convergence issues.
BrendanH
#9
May15-08, 10:37 PM
P: 63
Besides, we're dealing with polynomials in the case of (a+b)^n
Dick
#10
May15-08, 10:46 PM
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Quote Quote by BrendanH View Post
Besides, we're dealing with polynomials in the case of (a+b)^n
Yeah, that's what I mean by finite series. You could also just forget about pascal's triangle and multiply it out. The power is only 3.


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