Question 1179209
<pre>
The other tutor did not understand.  

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

Multiply by the LCD

{{{3x^2+1}}}{{{""=""}}}{{{A(x^""+2)(x^2+2x+2)}}}{{{""+""}}}{{{B(x^2+2x+2)}}}{{{""+""}}}{{{(Cx^""+D)(x^""+2)^2}}}

Since this must be an identity, we can substitute any numbers for x.
Substitute x=-2  (I chose -2 to make some terms become 0) 

{{{3(-2)^2+1}}}{{{""=""}}}{{{A((-2)^""+2)((-2)^2+2(-2)+2)}}}{{{""+""}}}{{{B((-2)^2+2(-2)+2)}}}{{{""+""}}}{{{(C(-2)^""+D)((-2)^""+2)^2}}}
{{{13}}}{{{""=""}}}{{{2B}}}
{{{13/2}}}{{{""=""}}}{{{B}}}

{{{3x^2+1}}}{{{""=""}}}{{{A(x^""+2)(x^2+2x+2)}}}{{{""+""}}}{{{(13/2)(x^2+2x+2)}}}{{{""+""}}}{{{(Cx^""+D)(x^""+2)^2}}}

{{{6x^2+2}}}{{{""=""}}}{{{2A(x^""+2)(x^2+2x+2)}}}{{{""+""}}}{{{13(x^2+2x+2)}}}{{{""+""}}}{{{2(Cx^""+D)(x^""+2)^2}}}

Substitute x=0  (I chose that because it's easy to substitute)

{{{6(0)^2+2}}}{{{""=""}}}{{{2A((0)^""+2)((0)^2+2(0)+2)}}}{{{""+""}}}{{{13((0)^2+2(0)+2)}}}{{{""+""}}}{{{2(C(0)^""+D)((0)^""+2)^2}}}

{{{2}}}{{{""=""}}}{{{2A(2)(2))}}}{{{""+""}}}{{{13(2)}}}{{{""+""}}}{{{2(D)(2)^2}}}

{{{2}}}{{{""=""}}}{{{8A}}}{{{""+""}}}{{{26}}}{{{""+""}}}{{{8D}}}

{{{-24}}}{{{""=""}}}{{{8A}}}{{{""+""}}}{{{8D}}}

{{{-3}}}{{{""=""}}}{{{A}}}{{{""+""}}}{{{D}}}

Substitute D = -A - 3

{{{6x^2+2}}}{{{""=""}}}{{{2A(x^""+2)(x^2+2x+2)}}}{{{""+""}}}{{{13(x^2+2x+2)}}}{{{""+""}}}{{{2(Cx^""-A-3)(x^""+2)^2}}}

Substitute x=1  

{{{6(1)^2+2}}}{{{""=""}}}{{{2A((1)^""+2)((1)^2+2(1)+2)}}}{{{""+""}}}{{{13((1)^2+2(1)+2)}}}{{{""+""}}}{{{2(C(1)^""-A-3)((1)^""+2)^2}}}

{{{8}}}{{{""=""}}}{{{2A(3)(5)}}}{{{""+""}}}{{{13(5)}}}{{{""+""}}}{{{2(C-A-3)(3)^2}}}

{{{8}}}{{{""=""}}}{{{30A}}}{{{""+""}}}{{{65}}}{{{""+""}}}{{{2(C-A-3)(9)}}}
{{{8}}}{{{""=""}}}{{{30A}}}{{{""+""}}}{{{65}}}{{{""+""}}}{{{18(C-A-3)}}}
{{{8}}}{{{""=""}}}{{{30A}}}{{{""+""}}}{{{65}}}{{{""+""}}}{{{18C-18A-54)}}}
{{{8}}}{{{""=""}}}{{{12A}}}{{{""+""}}}{{{11}}}{{{""+""}}}{{{18C)}}}
{{{8}}}{{{""=""}}}{{{12A}}}{{{""+""}}}{{{11}}}{{{""+""}}}{{{18C)}}}
{{{-3}}}{{{""=""}}}{{{12A}}}{{{""+""}}}{{{18C)}}}
{{{-1}}}{{{""=""}}}{{{4A}}}{{{""+""}}}{{{6C)}}}

Substitute x=-1  

{{{6(-1)^2+2}}}{{{""=""}}}{{{2A((-1)^""+2)((-1)^2+2(-1)+2)}}}{{{""+""}}}{{{13((-1)^2+2(-1)+2)}}}{{{""+""}}}{{{2(C(-1)^""-A-3)((-1)^""+2)^2}}}

{{{8}}}{{{""=""}}}{{{2A(1)(1)}}}{{{""+""}}}{{{13(1)}}}{{{""+""}}}{{{2(-C-A-3)(1)^2}}}
{{{8}}}{{{""=""}}}{{{2A}}}{{{""+""}}}{{{13}}}{{{""+""}}}{{{2(-C-A-3)}}}
{{{8}}}{{{""=""}}}{{{2A}}}{{{""+""}}}{{{13}}}{{{""-""}}}{{{2C-2A-6)}}}
{{{8}}}{{{""=""}}}{{{-2C}}}{{{""+""}}}{{{7}}}
{{{1}}}{{{""=""}}}{{{-2C}}}
{{{-1/2}}}{{{""=""}}}{{{C}}}

Substitute C = -1/2 in

{{{-1}}}{{{""=""}}}{{{4A}}}{{{""+""}}}{{{6C)}}}
{{{-1}}}{{{""=""}}}{{{4A}}}{{{""+""}}}{{{6(-1/2))}}}
{{{-1}}}{{{""=""}}}{{{4A}}}{{{""-""}}}{{{3)}}}
{{{2}}}{{{""=""}}}{{{4A}}}
{{{2/4}}}{{{""=""}}}{{{A}}}
{{{1/2}}}{{{""=""}}}{{{A}}}

Substitute A = 1/2 in 

{{{D}}}{{{""=""}}}{{{-A - 3}}}
{{{D}}}{{{""=""}}}{{{-1/2 - 3}}}
{{{D}}}{{{""=""}}}{{{-1/2 - 6/2}}}
{{{D}}}{{{""=""}}}{{{-7/2}}}

Substituting all the capital letters we've found:

{{{(3x^2+1)/((x-2^"")^2(x^2+2x+2))}}}{{{""=""}}}{{{1^""/(2(x^""+2^""^""))}}}{{{""+""}}}{{{13^""/(2(x^""+2)^2)}}}{{{""+""}}}{{{((-1/2)x^""^""+(-7/2))/(x^2^""+2x+2)}}}
 
So the final answer is:

{{{(3x^2+1)/((x-2^"")^2(x^2+2x+2))}}}{{{""=""}}}{{{1^""/(2(x^""+2^""))}}}{{{""+""}}}{{{13^""/(2(x^""+2)^2)}}}{{{""-""}}}{{{(x^""^""+7)/(2(x^2+2x+2))}}}

Edwin</pre>