<PRE><B>Question 1179209</B>
&lt;br&gt;
The first tutor misunderstood what was to be done.&lt;br&gt;
The second tutor rewrote the equation wrong so worked a different problem.&lt;br&gt;
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The fraction to be decomposed is&lt;br&gt;
{{{(3x^2+1)/((x-2)^2(x^2+2x+2))}}}&lt;br&gt;
From the basic rules for decomposing fractions, the decomposition will be of the form&lt;br&gt;
{{{(3x^2+1)/((x-2)^2(x^2+2x+2)) = A/(x-2)+B/(x-2)^2+(Cx+D)/(x^2+2x+2)}}}&lt;br&gt;
The standard procedure for doing the decomposition is to multiply the equation by the LCD:&lt;br&gt;
{{{3x^2+1 = A(x-2)(x^2+2x+2)+B(x^2+2x+2)+(Cx+D)(x-2)^2 = A(x-2)(x^2+2x+2)+B(x^2+2x+2)+(Cx+D)(x^2-4x+4)}}}&lt;br&gt;
Note that the second tutor in his work substituted &quot;convenient&quot; values for x to get equations in A, B, C, and D.  That turns out to give a quick path to the solution in many problems like this; but it doesn&#39;t appear to me to be advantageous in this example.&lt;br&gt;
So another standard method for getting equations in A, B, C, and D is to expand the expression on the right and equate coefficients on the two sides of the equation.&lt;br&gt;
{{{3x^2+1 = A(x^3+2x^2+2x-2x^2-4x-4)+B(x^2+2x+2)+(Cx^3-4Cx^2+4Cx+Dx^2-4Dx+4D)}}}
{{{3x^2+1 = (A+C)x^3+(B-4C+D)x^2+(-2A+2B+4C-4D)x+(-4A+2B+4D)}}}&lt;br&gt;
Equating the coefficients of the different degree terms on the two sides of the equation gives us four equations in A, B, C, and D:&lt;br&gt;
(1) A+C=0
(2) B-4C+D=3
(3) -2A+2B+4C-4D=0
(4) -4A+2B+4D=1&lt;br&gt;
Solve (1) to get C=-A and substitute in (2), (3), and (4):&lt;br&gt;
(5) 4A+B+D=3
(6) -6A+2B-4D=0
(7) -4A+2B+4D=1&lt;br&gt;
Eliminate D between (5) and (6), and between (6) and (7):&lt;br&gt;
(8) 10A+6B=12
(9) -10A+4B=1&lt;br&gt;
Add those two equations to eliminate A and solve for B:&lt;br&gt;
(10) 10B=13; B=13/10&lt;br&gt;
Substitute (10) in (8) and solve for A:&lt;br&gt;
10A+78/10=12
100A+78=120
100A=42
(11) A = 42/100 = 21/50&lt;br&gt;
Use (1) and (11) to get C = -21/50&lt;br&gt;
Use (5), (10), and (11) to find D:&lt;br&gt;
84/50+65/50+D=3
D=1/50&lt;br&gt;
We have values for all the unknowns:
A=21/50
B=13/10
C=-21/50
D=1/50&lt;br&gt;
ANSWER:&lt;br&gt;
{{{(3x^2+1)/((x-2)^2(x^2+2x+2)) = 21/(50(x-2))+13/(10(x-2)^2)+(-21x+1)/50(x^2+2x+2)}}}&lt;br&gt;</PRE>