Partial Fraction Expansion  Linear Factors (Single and Repeating)
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Single Linear Factors
When your factor is a linear factor, i.e. \(ax+b\), your numerator is always a constant. A linear factor means the highest power of the variable (\(x\) in this case) is one and \(a \neq 0\). Notice, we do not require that \(b\) be nonzero. So the term \(ax\) also falls in this category.
An example is \(\displaystyle{ \frac{1}{(x+1)(x+2)} }\), which has two linear factors, \((x+1)\) and \((x+2)\). Both of these factors are single, meaning that they each have a power \(1\), i.e. \((x+1) = (x+1)^1\). So we set up the partial fractions as
\(\displaystyle{ \frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} }\)
Repeated Linear Factors
What happens when we have a repeated linear factor? A repeating factor means that we have something like \((ax+b)^2\). This case has to be handled differently. We still use constants in the numerator (since we are still working with linear factors), but we need extra fractions. Let's set up an example to see what we need to do.
Example
\(\displaystyle{ \frac{1}{x(x+1)^2} = \frac{A_1}{x} + \frac{A_2}{x+1} + \frac{A_3}{(x+1)^2} }\)
Notice that we still use constants in the numerator because we have only linear factors (one of which is repeated) in the denominator. Then, we need two terms for the repeated term, one with power one, the second with power two.
In general, it looks like this for linear factors.
factor in the denominator 
partial fraction terms  

\(ax+b\) 
\(\displaystyle{ \frac{A}{ax+b} }\)  
\((ax+b)^2\) 
\(\displaystyle{ \frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} }\)  
\((ax+b)^3\) 
\(\displaystyle{ \frac{A_1}{ax+b} + }\) \(\displaystyle{\frac{A_2}{(ax+b)^2} + }\) \(\displaystyle{\frac{A_3}{(ax+b)^3} }\) 
Do you see the pattern? For a denominator with the term \((ax+b)^k\), we would have the factors
\(\displaystyle{ \frac{A_1}{ax+b} + }\)
\(\displaystyle{ \frac{A_2}{(ax+b)^2} + }\)
\(\displaystyle{ \frac{A_3}{(ax+b)^3} + . . . + } \)
\(\displaystyle{ \frac{A_k}{(ax+b)^k} }\)
Once these expansions are set up, the steps to find the constants are the same as shown in step 5 on the previous page.
Okay, time for the practice problems.
Practice
Unless otherwise instructed, expand these fractions using partial fraction expansion. Give your answers in exact terms.
\(\displaystyle{\frac{1}{y(1y)}}\)
Problem Statement 

Expand \(\displaystyle{\frac{1}{y(1y)}}\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{\frac{1}{y(1y)}}\) \(\displaystyle{ = \frac{1}{y} + \frac{1}{1y} }\)
Problem Statement
Expand \(\displaystyle{\frac{1}{y(1y)}}\) using partial fraction expansion. Give your answer in exact terms.
Solution
\(\displaystyle{\frac{1}{y(1y)}}\) \(\displaystyle{ = \frac{A}{y} + \frac{B}{1y} \to }\) \( 1 = A(1y) + B(y) \)
Choose \(y=0\)
\(1 = A(1y) + B(y) \to \) \( 1 = A(10) + B(0) \to \) \( A = 1 \)
Choose \(y=1\)
\(1 = A(1y) + B(y) \to \) \( 1 = A(11) + B(1) \to \) \( B = 1 \)
Final Answer
\(\displaystyle{\frac{1}{y(1y)}}\) \(\displaystyle{ = \frac{1}{y} + \frac{1}{1y} }\)
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\(\displaystyle{\frac{1}{(s5)(s2)}}\)
Problem Statement 

Expand \(\displaystyle{\frac{1}{(s5)(s2)}}\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{\frac{1}{(s5)(s2)}}\) \(\displaystyle{ = \frac{1/3}{s5} + \frac{1/3}{s2} }\)
Problem Statement
Expand \(\displaystyle{\frac{1}{(s5)(s2)}}\) using partial fraction expansion. Give your answer in exact terms.
Solution
\(\displaystyle{ \frac{1}{(s5)(s2)} = \frac{A}{s5} + \frac{B}{s2} }\) 
multiply both sides by \((s5)(s2)\) 
\(\displaystyle{ 1 = A(s2) + B(s5) }\) 
Let \( s=2 \) 
\( 1 = B(25) \to B=1/3 \) 
Let \( s=5 \) 
\( 1 = A(52) \to A=1/3 \) 
Final Answer
\(\displaystyle{\frac{1}{(s5)(s2)}}\) \(\displaystyle{ = \frac{1/3}{s5} + \frac{1/3}{s2} }\)
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\(\displaystyle{ \frac{7x23}{x^27x+10} }\)
Problem Statement 

Expand \(\displaystyle{ \frac{7x23}{x^27x+10} }\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{ \frac{7x23}{x^27x+10} }\)
Problem Statement
Expand \(\displaystyle{ \frac{7x23}{x^27x+10} }\) using partial fraction expansion. Give your answer in exact terms.
Solution
After he finds the answer he shows how to check his answer. We recommend that you check your answer as often as you have time for, especially on exams.
video by The Organic Chemistry Tutor 

Final Answer
\(\displaystyle{ \frac{7x23}{x^27x+10} }\)
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\(\displaystyle{ \frac{ 293x }{ x^2x6 } }\)
Problem Statement
Expand \(\displaystyle{ \frac{ 293x }{ x^2x6 } }\) using partial fraction expansion. Give your answer in exact terms.
Solution
video by The Organic Chemistry Tutor 

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\(\displaystyle{ \frac{ 19x^215x6 }{ 2x^33x^22x } }\)
Problem Statement
Expand \(\displaystyle{ \frac{ 19x^215x6 }{ 2x^33x^22x } }\) using partial fraction expansion.
Solution
video by The Organic Chemistry Tutor 

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\(\displaystyle{ \frac{ x^2+2x+18 }{ x(x3)^2 } }\)
Problem Statement
Expand \(\displaystyle{ \frac{ x^2+2x+18 }{ x(x3)^2 } }\) using partial fraction expansion. Give your answer in exact terms.
Solution
video by The Organic Chemistry Tutor 

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\(\displaystyle{ \frac{ 3x^26x20 }{ x^2(x5) } }\)
Problem Statement
Expand \(\displaystyle{ \frac{ 3x^26x20 }{ x^2(x5) } }\) using partial fraction expansion. Give your answer in exact terms.
Solution
video by The Organic Chemistry Tutor 

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\(\displaystyle{ \frac{ 5x3 }{ x^23x4 } }\)
Problem Statement 

Expand \(\displaystyle{ \frac{ 5x3 }{ x^23x4 } }\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{ \frac{ 5x3 }{ x^23x4 } }\) \(\displaystyle{ = \frac{1}{5} \left[ \frac{ 17 }{ x4 } + \frac{ 8 }{ x+1 } \right] }\)
Problem Statement
Expand \(\displaystyle{ \frac{ 5x3 }{ x^23x4 } }\) using partial fraction expansion. Give your answer in exact terms.
Solution
Final Answer
\(\displaystyle{ \frac{ 5x3 }{ x^23x4 } }\) \(\displaystyle{ = \frac{1}{5} \left[ \frac{ 17 }{ x4 } + \frac{ 8 }{ x+1 } \right] }\)
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\(\displaystyle{ \frac{ 6x22 }{ 2x^2+7x15 } }\)
Problem Statement 

Expand \(\displaystyle{ \frac{ 6x22 }{ 2x^2+7x15 } }\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{ \frac{ 6x22 }{ 2x^2+7x15 } }\) \(\displaystyle{ = \frac{ 4 }{ x+5 }  \frac{ 2 }{ 2x3 } }\)
Problem Statement
Expand \(\displaystyle{ \frac{ 6x22 }{ 2x^2+7x15 } }\) using partial fraction expansion. Give your answer in exact terms.
Solution
Final Answer
\(\displaystyle{ \frac{ 6x22 }{ 2x^2+7x15 } }\) \(\displaystyle{ = \frac{ 4 }{ x+5 }  \frac{ 2 }{ 2x3 } }\)
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\(\displaystyle{ \frac{ 7x11 }{ (x2)^2 } }\)
Problem Statement 

Expand \(\displaystyle{ \frac{ 7x11 }{ (x2)^2 } }\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{ \frac{ 7x11 }{ (x2)^2 } }\) \(\displaystyle{ = \frac{ 7 }{ x2 } + \frac{ 3 }{ (x2)^2 } }\)
Problem Statement
Expand \(\displaystyle{ \frac{ 7x11 }{ (x2)^2 } }\) using partial fraction expansion. Give your answer in exact terms.
Solution
Final Answer
\(\displaystyle{ \frac{ 7x11 }{ (x2)^2 } }\) \(\displaystyle{ = \frac{ 7 }{ x2 } + \frac{ 3 }{ (x2)^2 } }\)
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\(\displaystyle{ \frac{ 3x^224x+53 }{ x^36x^2+9x } }\)
Problem Statement 

Expand \(\displaystyle{ \frac{ 3x^224x+53 }{ x^36x^2+9x } }\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{ \frac{ 3x^224x+53 }{ x^36x^2+9x } }\) \(\displaystyle{ = \frac{ 53/9 }{ x } + \frac{ 5/2 }{ x3 } + \frac{ 8/3 }{ (x3)^2 } }\)
Problem Statement
Expand \(\displaystyle{ \frac{ 3x^224x+53 }{ x^36x^2+9x } }\) using partial fraction expansion. Give your answer in exact terms.
Solution
He initially writes the problem with \(53x\) instead of just \(53\) in the numerator. He later corrects this.
He does not finish the problem in this video, so here is our work to finish it out. His last equation is
\(\displaystyle{ 32 = \frac{ 212 }{ 9 }  2B + \frac{8}{3} }\)
\(\begin{array}{rcl} 2B & = & \displaystyle{ \frac{ 212 }{ 9 } + \frac{8}{3}  32 } \\ 2B & = & \displaystyle{ \frac{ 212 }{ 9 } + \frac{24}{9}  \frac{288}{9} } \\ 2B & = & \displaystyle{ \frac{ 45 }{ 9 } = 5 } \\ B & = & 5/2 \end{array} \)
Final Answer
\(\displaystyle{ \frac{ 3x^224x+53 }{ x^36x^2+9x } }\) \(\displaystyle{ = \frac{ 53/9 }{ x } + \frac{ 5/2 }{ x3 } + \frac{ 8/3 }{ (x3)^2 } }\)
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\(\displaystyle{ \frac{5x1}{x^2x2}} \)
Problem Statement
Expand \(\displaystyle{ \frac{5x1}{x^2x2}} \) using partial fraction expansion. Give your answer in exact terms.
Solution
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\(\displaystyle{ \frac{ x5 }{ (3x+5)(x2) } }\)
Problem Statement 

Expand \(\displaystyle{ \frac{ x5 }{ (3x+5)(x2) } }\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{ \frac{ x5 }{ (3x+5)(x2) } }\) \(\displaystyle{ = \frac{ 20/11 }{ 3x+5 } + \frac{ 3/11 }{ x2 } }\)
Problem Statement
Expand \(\displaystyle{ \frac{ x5 }{ (3x+5)(x2) } }\) using partial fraction expansion. Give your answer in exact terms.
Solution
Final Answer
\(\displaystyle{ \frac{ x5 }{ (3x+5)(x2) } }\) \(\displaystyle{ = \frac{ 20/11 }{ 3x+5 } + \frac{ 3/11 }{ x2 } }\)
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\(\displaystyle{ \frac{ 4x+1 }{ x(x+1)^2 } }\)
Problem Statement 

Expand \(\displaystyle{ \frac{ 4x+1 }{ x(x+1)^2 } }\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{ \frac{ 4x+1 }{ x(x+1)^2 } }\) \(\displaystyle{ = \frac{ 1 }{ x }  \frac{ 1 }{ x+1 } + \frac{ 3 }{ (x+1)^2 } }\)
Problem Statement
Expand \(\displaystyle{ \frac{ 4x+1 }{ x(x+1)^2 } }\) using partial fraction expansion. Give your answer in exact terms.
Solution
Final Answer
\(\displaystyle{ \frac{ 4x+1 }{ x(x+1)^2 } }\) \(\displaystyle{ = \frac{ 1 }{ x }  \frac{ 1 }{ x+1 } + \frac{ 3 }{ (x+1)^2 } }\)
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\(\displaystyle{ \frac{ 5x^2+11x+2 }{ x(x+1)^2 } }\)
Problem Statement 

Expand \(\displaystyle{ \frac{ 5x^2+11x+2 }{ x(x+1)^2 } }\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{ \frac{ 5x^2+11x+2 }{ x(x+1)^2 } }\) \(\displaystyle{ = \frac{ 2 }{ x } + \frac{ 3 }{ x+1 } + \frac{ 4 }{ (x+1)^2 } }\)
Problem Statement
Expand \(\displaystyle{ \frac{ 5x^2+11x+2 }{ x(x+1)^2 } }\) using partial fraction expansion. Give your answer in exact terms.
Solution
Final Answer
\(\displaystyle{ \frac{ 5x^2+11x+2 }{ x(x+1)^2 } }\) \(\displaystyle{ = \frac{ 2 }{ x } + \frac{ 3 }{ x+1 } + \frac{ 4 }{ (x+1)^2 } }\)
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\(\displaystyle{ \frac{ 3 }{ x^2+x2 } }\)
Problem Statement 

Expand \(\displaystyle{ \frac{ 3 }{ x^2+x2 } }\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{ \frac{ 3 }{ x^2+x2 } }\) \(\displaystyle{ = \frac{ 1 }{ x+2 } + \frac{ 1 }{ x1 } }\)
Problem Statement
Expand \(\displaystyle{ \frac{ 3 }{ x^2+x2 } }\) using partial fraction expansion. Give your answer in exact terms.
Solution
Final Answer
\(\displaystyle{ \frac{ 3 }{ x^2+x2 } }\) \(\displaystyle{ = \frac{ 1 }{ x+2 } + \frac{ 1 }{ x1 } }\)
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\(\displaystyle{ \frac{ 2x+2 }{ (x1)^2 } }\)
Problem Statement 

Expand \(\displaystyle{ \frac{ 2x+2 }{ (x1)^2 } }\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{ \frac{ 2x+2 }{ (x1)^2 } }\) \(\displaystyle{ = \frac{ 2 }{ (x1) } + \frac{ 4 }{ (x1)^2 } }\)
Problem Statement
Expand \(\displaystyle{ \frac{ 2x+2 }{ (x1)^2 } }\) using partial fraction expansion. Give your answer in exact terms.
Solution
Final Answer
\(\displaystyle{ \frac{ 2x+2 }{ (x1)^2 } }\) \(\displaystyle{ = \frac{ 2 }{ (x1) } + \frac{ 4 }{ (x1)^2 } }\)
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\(\displaystyle{ \frac{ x+4 }{ (x+1)^2 } }\)
Problem Statement 

Expand \(\displaystyle{ \frac{ x+4 }{ (x+1)^2 } }\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{ \frac{ x+4 }{ (x+1)^2 } }\) \( \displaystyle{ = \frac{ 1 }{ x+1 } + \frac{ 3 }{ (x+1)^2 } }\)
Problem Statement
Expand \(\displaystyle{ \frac{ x+4 }{ (x+1)^2 } }\) using partial fraction expansion. Give your answer in exact terms.
Solution
Final Answer
\(\displaystyle{ \frac{ x+4 }{ (x+1)^2 } }\) \( \displaystyle{ = \frac{ 1 }{ x+1 } + \frac{ 3 }{ (x+1)^2 } }\)
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\(\displaystyle{\frac{3x8}{x^24x5}}\)
Problem Statement
Expand \(\displaystyle{\frac{3x8}{x^24x5}}\) using partial fraction expansion. Give your answer in exact terms.
Solution
video by PatrickJMT 

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\(\displaystyle{\frac{4x+9}{(x1)(x+1)(x+4)}}\)
Problem Statement
Expand \(\displaystyle{\frac{4x+9}{(x1)(x+1)(x+4)}}\) using partial fraction expansion. Give your answer in exact terms.
Solution
video by PatrickJMT 

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\(\displaystyle{\frac{2x^2+4x5}{x^2(x+3)}}\)
Problem Statement
Expand \(\displaystyle{\frac{2x^2+4x5}{x^2(x+3)}}\) using partial fraction expansion. Give your answer in exact terms.
Solution
video by PatrickJMT 

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\(\displaystyle{\frac{12}{x^29}}\)
Problem Statement
Expand \(\displaystyle{\frac{12}{x^29}}\) using partial fraction expansion. Give your answer in exact terms.
Solution
video by MIP4U 

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\(\displaystyle{\frac{x+5}{x^2+4x+3}}\)
Problem Statement
Expand \(\displaystyle{\frac{x+5}{x^2+4x+3}}\) using partial fraction expansion. Give your answer in exact terms.
Solution
video by MIP4U 

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\(\displaystyle{\frac{5x^2+20x+6}{x^3+2x^2+x}}\)
Problem Statement
Expand \(\displaystyle{\frac{5x^2+20x+6}{x^3+2x^2+x}}\) using partial fraction expansion. Give your answer in exact terms.
Solution
video by MIP4U 

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\(\displaystyle{\frac{x^2+8}{x^3+4x^2}}\)
Problem Statement
Expand \(\displaystyle{\frac{x^2+8}{x^3+4x^2}}\) using partial fraction expansion. Give your answer in exact terms.
Solution
video by MIP4U 

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\(\displaystyle{\frac{x^2+2x5}{(x+1)(x^2+6x+9)}}\)
Problem Statement
Expand \(\displaystyle{\frac{x^2+2x5}{(x+1)(x^2+6x+9)}}\) using partial fraction expansion. Give your answer in exact terms.
Solution
video by PatrickJMT 

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\(\displaystyle{\frac{6x^219x+15}{(x1)(x2)^2}}\)
Problem Statement 

Expand \(\displaystyle{\frac{6x^219x+15}{(x1)(x2)^2}}\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{\frac{6x^219x+15}{(x1)(x2)^2}}\) \(\displaystyle{ = \frac{2}{x1} + \frac{4}{x2} + \frac{1}{(x2)^2} }\)
Problem Statement
Expand \(\displaystyle{\frac{6x^219x+15}{(x1)(x2)^2}}\) using partial fraction expansion. Give your answer in exact terms.
Solution
Set up the partial fraction expansion equation. 
\(\displaystyle{\frac{6x^219x+15}{(x1)(x2)^2}}\) \(\displaystyle{ = \frac{A}{x1} + \frac{B}{x2} + \frac{C}{(x2)^2} }\) 
Multiply both sides by the denominator \((x1)(x2)^2\) 
\( 6x^219x+15 \) \( = A(x2)^2 + B(x1)(x2) + C(x1)\) 
Now we can substitute various values for \(x\) and solve the equations. Any three, different values will do. However, if we choose some specific values, the equations will be easier to solve. We want to choose values that make as many of the terms on the right side of the equation equal to zero as possible (but not ALL terms equal to zero). So we will choose \(x=1\), \(x=2\) and \(x=0\). There first two choices make sense since two terms will turn out to be zero in each case. The third choice is so that we get a simpler expression on the left but any value will do, other than one and two, which we have already worked with. 
Let's start with \(x=1\). 
\(6  19 + 15\) \( = A(12)^2 + B(0) + C(0)\) \(\to A = 2\) 
Next, let \(x=2\). 
\( 6(2^2)  19(2) + 15 \) \( = A(0) + B(0) + C(21) \) \(\to C = 1\) 
Finally, let \(x=0\). 
\( 15 = A(02)^2 + B(01)(02) + C(01) \) 
Now, since we have already solved for \(A\) and \(C\), we can use those values in the last equation to solve for \(B\). 
\( 15 = 2(2)^2 + B(1)(2) + 1(1) \to B = 4 \) 
So our three unknowns are \( A=2, B=4, C=1 \) 
Another way to solve for \(A\), \(B\) and \(C\) is to multiply out the terms on the right side of \( 6x^219x+15 \) \( = A(x2)^2 + B(x1)(x2) + C(x1)\) and match coefficients. Sometimes, this is easier than what we did above. Let's try it. 
\( 6x^219x+15 \) \( = A(x2)^2 + B(x1)(x2) + C(x1)\) 
\( 6x^219x+15 \) \( = A(x^24x+4) + B(x^23x+2) + C(x1)\) 
\( 6x^219x+15 \) \( = Ax^24Ax+4A + Bx^23Bx+2B + CxC\) 
Now we group common terms on the right. 
\( 6x^219x+15 \) \( = (Ax^2 +Bx^2) + (4Ax 3Bx + Cx) +(4A +2B C)\) 
Now factor. 
\( 6x^219x+15 \) \( = (A +B)x^2 + (4A 3B + C)x +(4A +2B C)\) 
Next, we equate the coefficients on the left and right. 
For the \(x^2\) term, we have \( 6 = A+B \). 
For the \(x\) term, \(19 = 4A3B+C\) 
And finally, the constant, \(15 = 4A+2BC\) 
Now we have three equations and three unknowns. 
I am sure you know how to solve these. You can use substitution or reduction. This way did not turn out to be easier after all. 
Final Answer
\(\displaystyle{\frac{6x^219x+15}{(x1)(x2)^2}}\) \(\displaystyle{ = \frac{2}{x1} + \frac{4}{x2} + \frac{1}{(x2)^2} }\)
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\(\displaystyle{\frac{x^22x37}{x^23x40}}\)
Problem Statement 

Expand \(\displaystyle{\frac{x^22x37}{x^23x40}}\) using partial fraction expansion. Give your answer in exact terms.
Final Answer 

\(\displaystyle{\frac{x^22x37}{x^23x40}}\) \(\displaystyle{ = \frac{2/13}{x+5} + \frac{11/13}{x8} }\)
Problem Statement
Expand \(\displaystyle{\frac{x^22x37}{x^23x40}}\) using partial fraction expansion. Give your answer in exact terms.
Solution
Set up the partial fraction expansion equation. 
\(\displaystyle{\frac{x^22x37}{x^23x40}}\) \(\displaystyle{ = \frac{A}{x+5} + \frac{B}{x8} }\) 
Multiply both sides by the denominator \((x+5)(x8)\) 
\( x^22x37 \) \( = A(x8) + B(x+5) \) 
Now we can substitute various values for \(x\) and solve the equations. Any two, different values will do. However, if we choose some specific values, the equations will be easier to solve. We want to choose values that make as many of the terms on the right side of the equation equal to zero as possible (but not ALL terms equal to zero). So we will choose \(x=8\) and \(x=5\). 
\(8^2  2(8)  37\) \( = A(0) + B(8+5)\) \(\to B = 11/13\) 
Next, let \(x=5\). 
\( (5)^2 2(5)  37 \) \( = A(58) + B(0) \) \(\to A = 2/13\) 
So our two unknowns are \( A=2/13, B=11/13 \) 
Final Answer
\(\displaystyle{\frac{x^22x37}{x^23x40}}\) \(\displaystyle{ = \frac{2/13}{x+5} + \frac{11/13}{x8} }\)
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