Question 1133349
IN THE PARTIAL FRACTION DECOMPOSITION of x^2/(x-1)^2 (x+1)^2 the format is as follows:
{{{ x^2/((x-1)^2(x+1)^2) = A^""/(x-1)^"" + B^""/(x-1)^2 +C^""/(x+1)^"" + D^""/(x+1)^2}}}
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Multiply through by the LCD:

{{{ x^2 = A*(x-1)(x+1)^2 + B*(x+1)^2 +C*(x+1)(x-1)^2 + D*(x-1)^2}}}
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I let x = 1
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{{{ 1^2 = A*(1-1)(1+1)^2 + B*(1+1)^2 +C*(1+1)(1-1)^2 + D*(1-1)^2}}}

{{{ 1 = A*(0)(2)^2 + B*(2)^2 +C*(2)(0)^2 + D*(0)^2}}}

{{{ 1 = B*(2)^2}}}

{{{ 1 = B*4}}}

{{{1/4=B}}}
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and got B= 1/4 (which is correct) THEN let x=-1 to get D= 1/4
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{{{ (-1)^2 = A*(-1-1)(-1+1)^2 + B*(-1+1)^2 +C*(-1+1)(-1-1)^2 + D*(-1-1)^2}}}

{{{ 1 = A*(-2)(0)^2 + B*(0)^2 +C*(0)(-2)^2 + D*(-2)^2}}}

{{{ 1 = D*(-2)^2}}}

{{{ 1 = D*4}}}

{{{1/4=D}}}
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Am stuck at this point Can not get A or C
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Go back to:

{{{ x^2 = A*(x-1)(x+1)^2 + B*(x+1)^2 +C*(x+1)(x-1)^2 + D*(x-1)^2}}}

Substitute the values you got for B and D

{{{ x^2 = A*(x-1)(x+1)^2 + expr(1/4)(x+1)^2 +C*(x+1)(x-1)^2 + expr(1/4)(x-1)^2}}}

Clear of fractions:

{{{ 4*x^2 = 4A*(x-1)(x+1)^2 + (x+1)^2 +4C*(x+1)(x-1)^2 + (x-1)^2}}}

Substitute some number you haven't substituted before. 

(Note: You don't have to just substitute numbers that make terms equal to zero.
You can substitute ANY number for x and it will always give a true equation,
even if none of the terms become 0. So the easiest number you haven't
substituted is x=0.

{{{ 4*0^2 = 4A*(0-1)(0+1)^2 + (0+1)^2 +4C*(0+1)(0-1)^2 + (0-1)^2}}}
{{{ 0 = 4A*(-1)(1)^2 + (1)^2 +4C*(1)(-1)^2 + (-1)^2}}}
{{{ 0 = -4A + 1 +4C + 1}}}
{{{ 0 = -4A + 4C + 2}}}
{{{4A-4C=2}}}
{{{2A-2C=1}}}

The next easiest number you haven't substituted is x=2

{{{ 4*2^2 = 4A*(2-1)(2+1)^2 + (2+1)^2 +4C*(2+1)(2-1)^2 + (2-1)^2}}}
{{{ 4*4 = 4A*(1)(3)^2 + (3)^2 +4C*(3)(1)^2 + (1)^2}}}
{{{ 16 = 4A*(9) + 9 +4*C*(3) + 1}}}
{{{ 16 = 36A + 10 +12C}}}
{{{ 6 = 36A + 12C}}}
{{{ 1 = 6A + 2C}}}
{{{ 6A + 2C =1}}}

Solve the system of equations:

{{{system(2A-2C=1,6A + 2C =1)}}}

Get {{{A=1/4}}}, {{{C=-1/4}}}

That's how you do it. If you run out of numbers to substitute for x that
cause terms to be 0, substitute some numbers that DON'T cause any terms
to be 0, and solve a system of equations.  Any numbers will do.  I just
picked the next easiest numbers x=0 and x=2. 

Edwin</pre>