Question 1042243
<pre><b>
{{{(x^4 - x^2 + 2)/(x^3 - x^2)}}}

Put in zero placeholders for missing powers of x

{{{(x^4 +0x^3- x^2 +0x+ 2)/(x^3 - x^2+0x+0)}}}

          <u>             x+1</u>
x³-x²+0x+0)x&#8308;+0x³- x²+0x+2
           <u>x&#8308;- x³+0x²+0x</u>
               x³- x²+0x+2
               <u>x³- x²+0x+0</u>
                  0x²+0x+2

So 

{{{(x^4 - x^2 + 2)/(x^3 - x^2)}}}{{{""=""}}}{{{x+1+2/(x^3-x^2)}}}{{{""=""}}}

{{{x+1 + 2^""/(x^2(x-1))}}}{{{""=""}}}{{{x+1+A^""/x^""+B^""/x^2+C^""/(x-1)^""}}}

Just taking the fraction parts:

{{{2/(x^2(x-1))}}}{{{""=""}}}{{{A^""/x^""+B^""/x^2+C^""/(x-1)^""}}}

Multiply through by the LCD = x²(x-1)

{{{2}}}{{{""=""}}}{{{Ax(x-1)+B(x-1)+Cx^2}}}

If we let x=0, 2 terms on the right will become 0:

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

{{{2}}}{{{""=""}}}{{{-B}}}

{{{B}}}{{{""=""}}}{{{-2}}}, substituting:

{{{2}}}{{{""=""}}}{{{Ax(x-1)-2(x-1)+Cx^2}}}

If we let x=1, 2 terms on the right will become 0

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

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

{{{2}}}{{{""=""}}}{{{C}}}, substituting:

{{{2}}}{{{""=""}}}{{{Ax(x-1)-2(x-1)+2x^2}}}

Now we can let x be any number, say choose easy x=2

{{{2}}}{{{""=""}}}{{{A*2(2-1)-2(2-1)+2*2^2}}}

{{{2}}}{{{""=""}}}{{{A*2(1)-2(1)+2*4}}}

{{{2}}}{{{""=""}}}{{{2A-2+8}}}

{{{2}}}{{{""=""}}}{{{2A+6}}}

{{{-4}}}{{{""=""}}}{{{2A}}}

{{{-2}}}{{{""=""}}}{{{A}}}, 

substitute A=-2, B=-2, and C=2 in

{{{(x^4 - x^2 + 2)/(x^3 - x^2)}}}{{{""=""}}}{{{x+1 + (x^2+2)/(x^2(x-1))}}}{{{""=""}}}{{{x+1+A^""/x^""+B^""/x^2+C^""/(x-1)^""}}}{{{""=""}}}

{{{x+1-2^""/x^""-2^""/x^2+2^""/(x-1)^""}}}

Edwin</pre></b>