Algebra - Partial Fractions

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Section 5.5 : Partial Fractions

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5. Determine the partial fraction decomposition of each of the following expression.

$\frac{{6x + 5}}{{{{\left( {2x - 1} \right)}^2}}}$

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The first step is to determine the form of the partial fraction decomposition. For this problem the partial fraction decomposition is,

$\frac{{6x + 5}}{{{{\left( {2x - 1} \right)}^2}}} = \frac{A}{{2x - 1}} + \frac{B}{{{{\left( {2x - 1} \right)}^2}}}$ Show Step 2

The LCD for this expression is ${\left( {2x - 1} \right)^2}$ . Adding the terms back up gives,

$\frac{{6x + 5}}{{{{\left( {2x - 1} \right)}^2}}} = \frac{{A\left( {2x - 1} \right) + B}}{{{{\left( {2x - 1} \right)}^2}}}$ Show Step 3

Setting the numerators equal gives,

$6x + 5 = A\left( {2x - 1} \right) + B$ Show Step 4

For this problem we can pick a “good” value of $x$ to determine only one of the constants. Here is that work.

$x = \frac{1}{2}: \hspace{0.25in} 8 = B \hspace{0.25in} \to \hspace{0.25in}B = 8$ Show Step 5

To get the remaining constant we can use any value of $x$ and plug that along with the value of $B$ we found in the previous step into the equation from Step 3.

It really doesn’t matter what value of $x$ we pick as long as it isn’t $x = \frac{1}{2}$ since we used that in the previous step. The idea here is to pick a value of $x$ that won’t create “large” or “messy” numbers, if possible. Good choices are often $x = 0$ or $x = 1$ , provided they weren’t used in the previous step of course.

For this problem $x = 0$ seems to be a good choice. Here is the work for this step.

$\begin{align*}6\left( 0 \right) + 5 & = A\left( {2\left( 0 \right) - 1} \right) + 8\\ 5 & = - A + 8\\ - 3 & = - A\hspace{0.25in} \Rightarrow \hspace{0.25in}A = 3\end{align*}$ Show Step 6

The partial fraction decomposition is then,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{6x + 5}}{{{{\left( {2x - 1} \right)}^2}}} = \frac{3}{{2x - 1}} + \frac{8}{{{{\left( {2x - 1} \right)}^2}}}}}$