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Partial Fraction Expansions

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Partial Fractions

When given a rational expression like \frac{4x-9}{x^2-3x}  it is very helpful in calculus to be able to write it as the sum of two simpler fractions like \frac{3}{x}+\frac{1}{x-3} .  The challenging part is trying to get from the initial rational expression to the simpler fractions

You may know how to add fractions and go from two or more separate fractions to a single fraction, but how do you go the other way around? 

Watch This

http://www.youtube.com/watch?v=S-XKGBesRzk Khan Academy: Partial Fraction Expansion 1

Guidance

Partial fraction decomposition is a procedure that reverses adding fractions with unlike denominators.  The most challenging part is coming up with the denominators of each individual partial fraction.  See if you can spot the pattern.

\frac{6x-1}{x^2 (x-1)(x^2+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}+\frac{Dx+E}{x^2+2}

In this example each individual factor must be represented.  Linear factors that are raised to a power greater than one must have each successive power included as a separate denominator.  Quadratic terms that do not factor to be linear terms are included with a numerator that is a linear function of x .

Example A

Use partial fractions to decompose the following rational expression.

\frac{7x^2+x+6}{x^3+3x}

Solution:  First factor the denominator and identify the denominators of the partial fractions.

\frac{7x^2+x+6}{x(x^2+3)}=\frac{A}{x}+\frac{Bx+C}{x^2+3}

When the fractions are eliminated by multiplying through by the LCD the equation becomes:

7x^2+x+6 &= A(x^2+3)+x(Bx+C)\\7x^2+x+6 &= Ax^2+3A+Bx^2+Cx

Notice the squared term, linear term and constant term form a system of three equations with three variables.

A+B &= 7\\C &= 1\\3A &= 6

In this case it is easy to see that A=2, B=5, C=1 . Often, the resulting system of equations is more complex and would benefit from your knowledge of solving systems using matrices.

\frac{7x^2+x+6}{x(x^2+3)}=\frac{2}{x}+\frac{5x+1}{x^2+3}

Example B

Use matrices to complete the partial fraction decomposition of the following rational expression.

\frac{2x+4}{(x-1)(x+3)}

Solution:

\frac{2x+4}{(x-1)(x+3)} &= \frac{A}{x+1}+\frac{B}{x+3}\\2x+4 &= Ax+3A+Bx+B\\

2 &= A+B\\4 &= 3A+B\\

\begin{bmatrix}\begin{matrix}1 & 1\\3 & 1\end{matrix} \left| \begin{matrix} 2\\4\end{matrix}\right.\end{bmatrix} \begin{matrix}& \rightarrow &\\\rightarrow & -3R_1 + R_2 & \rightarrow\end{matrix} & \begin{bmatrix}\begin{matrix}1 & 1\\0 & -2\end{matrix} \left| \begin{matrix} 2\\-2\end{matrix}\right.\end{bmatrix} \begin{matrix}& \rightarrow &\\\rightarrow & \div -2 & \rightarrow\end{matrix} \begin{bmatrix}\begin{matrix}1 & 1\\0 & 1\end{matrix} \left| \begin{matrix} 2\\1\end{matrix}\right.\end{bmatrix} \begin{matrix}\rightarrow & -R_2 + R_1 & \rightarrow\\& \rightarrow &\end{matrix} \begin{bmatrix}\begin{matrix}1 & 0\\0 & 1\end{matrix} \left| \begin{matrix} 1\\1\end{matrix}\right.\end{bmatrix}\\

A=1, \ & B=1\\

\frac{2x+4}{(x-1)(x+3)} &= \frac{1}{x+1}+\frac{1}{x+3}

Vocabulary

Partial fraction decomposition is a procedure that undoes the operation of adding fractions with unlike denominators.  It separates a rational expression into the sum of rational expressions with unlike denominators. 

Guided Practice

1. Use matrices to help you decompose the following rational expression.

\frac{5x-2}{(2x-1)(3x+4)}

2. Confirm Example C by adding the partial fractions.

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

3. Confirm Guided Practice #1 by adding the partial fractions.

\frac{5x-2}{(2x-1)(3x+4)}=\frac{\frac{1}{11}}{2x-1}+\frac{-\frac{26}{11}}{3x+4}

Answers:

1. \frac{5x-2}{(2x-1)(3x+4)} &= \frac{A}{2x-1}+\frac{B}{3x+4}\\5x-2 &= A(3x+4)+B(2x-1)\\5x-2 &= 3Ax+4A+2Bx-B\\5 &= 3A+2B\\-2 &= 4A-B\\\begin{bmatrix}\begin{matrix}3 & 2\\4 & -1\end{matrix} \left| \begin{matrix}5\\-2\end{matrix}\right.\end{bmatrix} \begin{matrix}\rightarrow & \cdot 4 & \rightarrow\\\rightarrow & \cdot 3 & \rightarrow\end{matrix} & \begin{bmatrix}\begin{matrix}12 & 8\\12 & -3\end{matrix} \left| \begin{matrix}20\\-6\end{matrix}\right.\end{bmatrix} \begin{matrix}& \rightarrow &\\\rightarrow & -R_1 + R_2 & \rightarrow\end{matrix} \begin{bmatrix}\begin{matrix}12 & 8\\0 & -11\end{matrix} \left| \begin{matrix}20\\-26\end{matrix}\right.\end{bmatrix} \begin{matrix}\rightarrow & \cdot 11 & \rightarrow\\\rightarrow & \cdot 8 & \rightarrow\end{matrix} \begin{bmatrix}\begin{matrix}132 & 88\\0 & -88\end{matrix} \left| \begin{matrix}220\\-208\end{matrix}\right.\end{bmatrix} \\& \begin{matrix}\rightarrow & +R_2 + R_1 & \rightarrow\\& \rightarrow &\end{matrix} \begin{bmatrix}\begin{matrix}132 & 0\\0 & -88\end{matrix} \left| \begin{matrix}12\\-208\end{matrix}\right.\end{bmatrix}  \begin{matrix}\rightarrow & \div 132 & \rightarrow\\\rightarrow & \div -88 & \rightarrow\end{matrix} \begin{bmatrix}\begin{matrix}1 & 0\\0 & 1\end{matrix} \left| \begin{matrix}\frac{1}{11}\\\frac{26}{11}\end{matrix}\right.\end{bmatrix}\\A = \frac{1}{11}, \ & B =-\frac{26}{11}\\\frac{5x-2}{(2x-1)(3x+4)} &= \frac{\frac{1}{11}}{2x-1}+\frac{\frac{26}{11}}{3x+4}

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

3. \frac{5x-2}{(2x-1)(3x+4)} &= \frac{\frac{1}{11}}{2x-1}+\frac{\frac{26}{11}}{3x+4}\\5x-2 &= \frac{1}{11}(3x+4)+\frac{26}{11}(2x-1)\\55x-22 &= 3x+4+26(2x-1)\\55x-22 &= 3x+4+52x-26\\55x-22 &= 55x-22


Homework

Decompose the following rational expressions.  Practice using matrices with at least one of the problems.

1.  \frac{3x-4}{(x-1)(x+4)}

2.  \frac{2x+1}{x^2(x-3)}

3.  \frac{x+1}{x(x-5)}

4. \frac{x^2+3x+1}{x(x-3)(x+6)}

5.  \frac{x^2+1}{x(x-1)(x+1)}

6.  \frac{4x^2-9}{x^2(x-4)}

7.  \frac{2x-4}{(x+7)(x-3)}

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

9. Confirm your answer to #1 by adding the partial fractions.

10. Confirm your answer to #5 by adding the partial fractions.

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