- #1
kostoglotov
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Homework Statement
Decompose [itex]\frac{2(1-2x^2)}{x(1-x^2)}[/itex]
I get A = 2, B =-1, C = 1, but this doesn't recompose into the correct equation, and the calculators for partial fraction decomposition online all agree that it should be A = 2, B = 1, C = 1.
Here is one of the online calculator results with steps shown: http://www.emathhelp.net/calculators/algebra-2/partial-fraction-decomposition-calculator/?numer=2(1-2x^2)&denom=x(1-x)(1+x)&steps=on
Homework Equations
The Attempt at a Solution
[/B]
[tex]\frac{2(1-2x^2)}{x(1-x^2)} = \frac{2(1-2x^2)}{x(1-x)(1+x)}[/tex]
[tex]\frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x} = \frac{2(1-2x^2)}{x(1-x)(1+x)}[/tex]
[tex]A(1-x)(1+x) + Bx(1+x) + Cx(1-x) = 2(1-2x^2)[/tex]
Let (i) x = 0, (ii) x = +1, (iii) x = -1
[tex]x = 0, A = 2(1-0), A = 2[/tex]
[tex]x = 1, B(1)(1+(1)) = 2(1-2) = -2, 2B = -2, B = -1[/tex]
[tex]x = -1, C(-1)(1-(-1)) = 2(1-2) = -2, -2C = -2, C = 1 [/tex]
This of course recomposes to
[tex]\frac{2(1-x^2)-2x^2}{x(1-x)(1+x)}[/tex]
not the original equation.
The only thing the online calculator really did differently from me is change a term in the denominator from [itex]1-x^2[/itex] to [itex]x^2-1[/itex]. I used a shortcut of subbing in the roots of the denominator in order to quickly find the constant numerator values, but even if I use the long method of matching the term coefficients as the online calculator used, I still get my wrong values for A,B and C...why does changing [itex]1-x^2[/itex] to [itex]x^2-1[/itex] matter?
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