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17Calculus Precalculus - Linear Partial Fraction Expansion

17Calculus
Single Variable Calculus
Derivatives
Integrals
Multi-Variable Calculus
Precalculus
Functions

Partial Fraction Expansion - Linear Factors (Single and Repeating)

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 non-zero. 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.

Learn to Remember: Practical Techniques and Exercises to Improve Your Memory

Practice

Unless otherwise instructed, expand these fractions using partial fraction expansion. Give your answers in exact terms.

\(\displaystyle{\frac{1}{y(1-y)}}\)

Problem Statement

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

Final Answer

\(\displaystyle{\frac{1}{y(1-y)}}\) \(\displaystyle{ = \frac{1}{y} + \frac{1}{1-y} }\)

Problem Statement

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

Solution

\(\displaystyle{\frac{1}{y(1-y)}}\) \(\displaystyle{ = \frac{A}{y} + \frac{B}{1-y} \to }\) \( 1 = A(1-y) + B(y) \)

Choose \(y=0\)
\(1 = A(1-y) + B(y) \to \) \( 1 = A(1-0) + B(0) \to \) \( A = 1 \)

Choose \(y=1\)
\(1 = A(1-y) + B(y) \to \) \( 1 = A(1-1) + B(1) \to \) \( B = 1 \)

Final Answer

\(\displaystyle{\frac{1}{y(1-y)}}\) \(\displaystyle{ = \frac{1}{y} + \frac{1}{1-y} }\)

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\(\displaystyle{\frac{1}{(s-5)(s-2)}}\)

Problem Statement

Expand \(\displaystyle{\frac{1}{(s-5)(s-2)}}\) using partial fraction expansion. Give your answer in exact terms.

Final Answer

\(\displaystyle{\frac{1}{(s-5)(s-2)}}\) \(\displaystyle{ = \frac{1/3}{s-5} + \frac{-1/3}{s-2} }\)

Problem Statement

Expand \(\displaystyle{\frac{1}{(s-5)(s-2)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

\(\displaystyle{ \frac{1}{(s-5)(s-2)} = \frac{A}{s-5} + \frac{B}{s-2} }\)

multiply both sides by \((s-5)(s-2)\)

\(\displaystyle{ 1 = A(s-2) + B(s-5) }\)

Let \( s=2 \)

\( 1 = B(2-5) \to B=-1/3 \)

Let \( s=5 \)

\( 1 = A(5-2) \to A=1/3 \)

Final Answer

\(\displaystyle{\frac{1}{(s-5)(s-2)}}\) \(\displaystyle{ = \frac{1/3}{s-5} + \frac{-1/3}{s-2} }\)

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\(\displaystyle{ \frac{7x-23}{x^2-7x+10} }\)

Problem Statement

Expand \(\displaystyle{ \frac{7x-23}{x^2-7x+10} }\) using partial fraction expansion. Give your answer in exact terms.

Final Answer

\(\displaystyle{ \frac{7x-23}{x^2-7x+10} }\)

Problem Statement

Expand \(\displaystyle{ \frac{7x-23}{x^2-7x+10} }\) 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.

The Organic Chemistry Tutor - 3083 video solution

Final Answer

\(\displaystyle{ \frac{7x-23}{x^2-7x+10} }\)

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\(\displaystyle{ \frac{ 29-3x }{ x^2-x-6 } }\)

Problem Statement

Expand \(\displaystyle{ \frac{ 29-3x }{ x^2-x-6 } }\) using partial fraction expansion. Give your answer in exact terms.

Solution

The Organic Chemistry Tutor - 3084 video solution

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\(\displaystyle{ \frac{ 19x^2-15x-6 }{ 2x^3-3x^2-2x } }\)

Problem Statement

Expand \(\displaystyle{ \frac{ 19x^2-15x-6 }{ 2x^3-3x^2-2x } }\) using partial fraction expansion.

Solution

The Organic Chemistry Tutor - 3085 video solution

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\(\displaystyle{ \frac{ -x^2+2x+18 }{ x(x-3)^2 } }\)

Problem Statement

Expand \(\displaystyle{ \frac{ -x^2+2x+18 }{ x(x-3)^2 } }\) using partial fraction expansion. Give your answer in exact terms.

Solution

The Organic Chemistry Tutor - 3086 video solution

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\(\displaystyle{ \frac{ -3x^2-6x-20 }{ x^2(x-5) } }\)

Problem Statement

Expand \(\displaystyle{ \frac{ -3x^2-6x-20 }{ x^2(x-5) } }\) using partial fraction expansion. Give your answer in exact terms.

Solution

The Organic Chemistry Tutor - 3087 video solution

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\(\displaystyle{ \frac{ 5x-3 }{ x^2-3x-4 } }\)

Problem Statement

Expand \(\displaystyle{ \frac{ 5x-3 }{ x^2-3x-4 } }\) using partial fraction expansion. Give your answer in exact terms.

Final Answer

\(\displaystyle{ \frac{ 5x-3 }{ x^2-3x-4 } }\) \(\displaystyle{ = \frac{1}{5} \left[ \frac{ 17 }{ x-4 } + \frac{ 8 }{ x+1 } \right] }\)

Problem Statement

Expand \(\displaystyle{ \frac{ 5x-3 }{ x^2-3x-4 } }\) using partial fraction expansion. Give your answer in exact terms.

Solution

3088 video solution

Final Answer

\(\displaystyle{ \frac{ 5x-3 }{ x^2-3x-4 } }\) \(\displaystyle{ = \frac{1}{5} \left[ \frac{ 17 }{ x-4 } + \frac{ 8 }{ x+1 } \right] }\)

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\(\displaystyle{ \frac{ 6x-22 }{ 2x^2+7x-15 } }\)

Problem Statement

Expand \(\displaystyle{ \frac{ 6x-22 }{ 2x^2+7x-15 } }\) using partial fraction expansion. Give your answer in exact terms.

Final Answer

\(\displaystyle{ \frac{ 6x-22 }{ 2x^2+7x-15 } }\) \(\displaystyle{ = \frac{ 4 }{ x+5 } - \frac{ 2 }{ 2x-3 } }\)

Problem Statement

Expand \(\displaystyle{ \frac{ 6x-22 }{ 2x^2+7x-15 } }\) using partial fraction expansion. Give your answer in exact terms.

Solution

3089 video solution

Final Answer

\(\displaystyle{ \frac{ 6x-22 }{ 2x^2+7x-15 } }\) \(\displaystyle{ = \frac{ 4 }{ x+5 } - \frac{ 2 }{ 2x-3 } }\)

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\(\displaystyle{ \frac{ 7x-11 }{ (x-2)^2 } }\)

Problem Statement

Expand \(\displaystyle{ \frac{ 7x-11 }{ (x-2)^2 } }\) using partial fraction expansion. Give your answer in exact terms.

Final Answer

\(\displaystyle{ \frac{ 7x-11 }{ (x-2)^2 } }\) \(\displaystyle{ = \frac{ 7 }{ x-2 } + \frac{ 3 }{ (x-2)^2 } }\)

Problem Statement

Expand \(\displaystyle{ \frac{ 7x-11 }{ (x-2)^2 } }\) using partial fraction expansion. Give your answer in exact terms.

Solution

3090 video solution

Final Answer

\(\displaystyle{ \frac{ 7x-11 }{ (x-2)^2 } }\) \(\displaystyle{ = \frac{ 7 }{ x-2 } + \frac{ 3 }{ (x-2)^2 } }\)

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\(\displaystyle{ \frac{ 3x^2-24x+53 }{ x^3-6x^2+9x } }\)

Problem Statement

Expand \(\displaystyle{ \frac{ 3x^2-24x+53 }{ x^3-6x^2+9x } }\) using partial fraction expansion. Give your answer in exact terms.

Final Answer

\(\displaystyle{ \frac{ 3x^2-24x+53 }{ x^3-6x^2+9x } }\) \(\displaystyle{ = \frac{ 53/9 }{ x } + \frac{ -5/2 }{ x-3 } + \frac{ 8/3 }{ (x-3)^2 } }\)

Problem Statement

Expand \(\displaystyle{ \frac{ 3x^2-24x+53 }{ x^3-6x^2+9x } }\) 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
\(\displaystyle{ 32 = \frac{ 212 }{ 9 } - 2B + \frac{8}{3} }\)
\(\begin{array}{rcl} 2B & = & \displaystyle{ \frac{ 212 }{ 9 } + \frac{8}{3} - 32 } \\ 2B & = & \displaystyle{ \frac{ 212 }{ 9 } + \frac{24}{9} - \frac{288}{9} } \\ 2B & = & \displaystyle{ \frac{ -45 }{ 9 } = -5 } \\ B & = & -5/2 \end{array} \)

3091 video solution

Final Answer

\(\displaystyle{ \frac{ 3x^2-24x+53 }{ x^3-6x^2+9x } }\) \(\displaystyle{ = \frac{ 53/9 }{ x } + \frac{ -5/2 }{ x-3 } + \frac{ 8/3 }{ (x-3)^2 } }\)

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\(\displaystyle{ \frac{5x-1}{x^2-x-2}} \)

Problem Statement

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

Solution

3092 video solution

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\(\displaystyle{ \frac{ x-5 }{ (3x+5)(x-2) } }\)

Problem Statement

Expand \(\displaystyle{ \frac{ x-5 }{ (3x+5)(x-2) } }\) using partial fraction expansion. Give your answer in exact terms.

Final Answer

\(\displaystyle{ \frac{ x-5 }{ (3x+5)(x-2) } }\) \(\displaystyle{ = \frac{ 20/11 }{ 3x+5 } + \frac{ -3/11 }{ x-2 } }\)

Problem Statement

Expand \(\displaystyle{ \frac{ x-5 }{ (3x+5)(x-2) } }\) using partial fraction expansion. Give your answer in exact terms.

Solution

3094 video solution

Final Answer

\(\displaystyle{ \frac{ x-5 }{ (3x+5)(x-2) } }\) \(\displaystyle{ = \frac{ 20/11 }{ 3x+5 } + \frac{ -3/11 }{ x-2 } }\)

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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

3095 video 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

3096 video 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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\(\displaystyle{ \frac{ 3 }{ x^2+x-2 } }\)

Problem Statement

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

Final Answer

\(\displaystyle{ \frac{ 3 }{ x^2+x-2 } }\) \(\displaystyle{ = \frac{ -1 }{ x+2 } + \frac{ 1 }{ x-1 } }\)

Problem Statement

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

Solution

3097 video solution

Final Answer

\(\displaystyle{ \frac{ 3 }{ x^2+x-2 } }\) \(\displaystyle{ = \frac{ -1 }{ x+2 } + \frac{ 1 }{ x-1 } }\)

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\(\displaystyle{ \frac{ 2x+2 }{ (x-1)^2 } }\)

Problem Statement

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

Final Answer

\(\displaystyle{ \frac{ 2x+2 }{ (x-1)^2 } }\) \(\displaystyle{ = \frac{ 2 }{ (x-1) } + \frac{ 4 }{ (x-1)^2 } }\)

Problem Statement

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

Solution

3098 video solution

Final Answer

\(\displaystyle{ \frac{ 2x+2 }{ (x-1)^2 } }\) \(\displaystyle{ = \frac{ 2 }{ (x-1) } + \frac{ 4 }{ (x-1)^2 } }\)

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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

3099 video solution

Final Answer

\(\displaystyle{ \frac{ x+4 }{ (x+1)^2 } }\) \( \displaystyle{ = \frac{ 1 }{ x+1 } + \frac{ 3 }{ (x+1)^2 } }\)

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\(\displaystyle{\frac{3x-8}{x^2-4x-5}}\)

Problem Statement

Expand \(\displaystyle{\frac{3x-8}{x^2-4x-5}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

PatrickJMT - 1483 video solution

video by PatrickJMT

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\(\displaystyle{\frac{4x+9}{(x-1)(x+1)(x+4)}}\)

Problem Statement

Expand \(\displaystyle{\frac{4x+9}{(x-1)(x+1)(x+4)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

PatrickJMT - 1484 video solution

video by PatrickJMT

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\(\displaystyle{\frac{2x^2+4x-5}{x^2(x+3)}}\)

Problem Statement

Expand \(\displaystyle{\frac{2x^2+4x-5}{x^2(x+3)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

PatrickJMT - 1485 video solution

video by PatrickJMT

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\(\displaystyle{\frac{12}{x^2-9}}\)

Problem Statement

Expand \(\displaystyle{\frac{12}{x^2-9}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

MIP4U - 1491 video 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

MIP4U - 1492 video 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

MIP4U - 1493 video 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

MIP4U - 1494 video solution

video by MIP4U

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\(\displaystyle{\frac{-x^2+2x-5}{(x+1)(x^2+6x+9)}}\)

Problem Statement

Expand \(\displaystyle{\frac{-x^2+2x-5}{(x+1)(x^2+6x+9)}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

PatrickJMT - 1486 video solution

video by PatrickJMT

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\(\displaystyle{\frac{6x^2-19x+15}{(x-1)(x-2)^2}}\)

Problem Statement

Expand \(\displaystyle{\frac{6x^2-19x+15}{(x-1)(x-2)^2}}\) using partial fraction expansion. Give your answer in exact terms.

Final Answer

\(\displaystyle{\frac{6x^2-19x+15}{(x-1)(x-2)^2}}\) \(\displaystyle{ = \frac{2}{x-1} + \frac{4}{x-2} + \frac{1}{(x-2)^2} }\)

Problem Statement

Expand \(\displaystyle{\frac{6x^2-19x+15}{(x-1)(x-2)^2}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

Set up the partial fraction expansion equation.

\(\displaystyle{\frac{6x^2-19x+15}{(x-1)(x-2)^2}}\) \(\displaystyle{ = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{(x-2)^2} }\)

Multiply both sides by the denominator \((x-1)(x-2)^2\)

\( 6x^2-19x+15 \) \( = A(x-2)^2 + B(x-1)(x-2) + C(x-1)\)

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 \(x=1\).

\(6 - 19 + 15\) \( = A(1-2)^2 + B(0) + C(0)\) \(\to A = 2\)

Next, let \(x=2\).

\( 6(2^2) - 19(2) + 15 \) \( = A(0) + B(0) + C(2-1) \) \(\to C = 1\)

Finally, let \(x=0\).

\( 15 = A(0-2)^2 + B(0-1)(0-2) + C(0-1) \)

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 \( 6x^2-19x+15 \) \( = A(x-2)^2 + B(x-1)(x-2) + C(x-1)\) and match coefficients. Sometimes, this is easier than what we did above. Let's try it.

\( 6x^2-19x+15 \) \( = A(x-2)^2 + B(x-1)(x-2) + C(x-1)\)

\( 6x^2-19x+15 \) \( = A(x^2-4x+4) + B(x^2-3x+2) + C(x-1)\)

\( 6x^2-19x+15 \) \( = Ax^2-4Ax+4A + Bx^2-3Bx+2B + Cx-C\)

Now we group common terms on the right.

\( 6x^2-19x+15 \) \( = (Ax^2 +Bx^2) + (-4Ax -3Bx + Cx) +(4A +2B -C)\)

Now factor.

\( 6x^2-19x+15 \) \( = (A +B)x^2 + (-4A -3B + C)x +(4A +2B -C)\)

Next, we equate the coefficients on the left and right.

For the \(x^2\) term, we have \( 6 = A+B \).

For the \(x\) term, \(-19 = -4A-3B+C\)

And finally, the constant, \(15 = 4A+2B-C\)

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

\(\displaystyle{\frac{6x^2-19x+15}{(x-1)(x-2)^2}}\) \(\displaystyle{ = \frac{2}{x-1} + \frac{4}{x-2} + \frac{1}{(x-2)^2} }\)

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\(\displaystyle{\frac{x^2-2x-37}{x^2-3x-40}}\)

Problem Statement

Expand \(\displaystyle{\frac{x^2-2x-37}{x^2-3x-40}}\) using partial fraction expansion. Give your answer in exact terms.

Final Answer

\(\displaystyle{\frac{x^2-2x-37}{x^2-3x-40}}\) \(\displaystyle{ = \frac{2/13}{x+5} + \frac{11/13}{x-8} }\)

Problem Statement

Expand \(\displaystyle{\frac{x^2-2x-37}{x^2-3x-40}}\) using partial fraction expansion. Give your answer in exact terms.

Solution

Set up the partial fraction expansion equation.

\(\displaystyle{\frac{x^2-2x-37}{x^2-3x-40}}\) \(\displaystyle{ = \frac{A}{x+5} + \frac{B}{x-8} }\)

Multiply both sides by the denominator \((x+5)(x-8)\)

\( x^2-2x-37 \) \( = A(x-8) + B(x+5) \)

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 \(x=-5\).

\( (-5)^2 -2(-5) - 37 \) \( = A(-5-8) + B(0) \) \(\to A = 2/13\)

So our two unknowns are \( A=2/13, B=11/13 \)

Final Answer

\(\displaystyle{\frac{x^2-2x-37}{x^2-3x-40}}\) \(\displaystyle{ = \frac{2/13}{x+5} + \frac{11/13}{x-8} }\)

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partial fraction expansion

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Practice Instructions

Unless otherwise instructed, expand these fractions using partial fraction expansion. Give your answers in exact terms.

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