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### Section 5.5 : Partial Fractions

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

$\frac{{4{x^2} - 22x + 7}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}}$

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Start Solution

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

$\frac{{4{x^2} - 22x + 7}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}} = \frac{A}{{2x + 3}} + \frac{B}{{x - 2}} + \frac{C}{{{{\left( {x - 2} \right)}^2}}}$ Show Step 2

The LCD for this expression is $$\left( {2x + 3} \right){\left( {x - 2} \right)^2}$$. Adding the terms back up gives,

$\frac{{4{x^2} - 22x + 7}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}} = \frac{{A{{\left( {x - 2} \right)}^2} + B\left( {2x + 3} \right)\left( {x - 2} \right) + C\left( {2x + 3} \right)}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}}$ Show Step 3

Setting the numerators equal gives,

$4{x^2} - 22x + 7 = A{\left( {x - 2} \right)^2} + B\left( {2x + 3} \right)\left( {x - 2} \right) + C\left( {2x + 3} \right)$ Show Step 4

For this problem we can pick “good” values of $$x$$ to determine only two of the three constants. Here is that work.

\begin{array}{l}x = - \frac{3}{2}:}\\{x = 2:}\end{array}\hspace{0.25in}\begin{aligned}49 & = A{{\left( { - \frac{7}{2}} \right)}^2} = \frac{{49}}{4}A\\ - 21 & = 7C\end{aligned}\hspace{0.25in} \to \hspace{0.25in}\begin{array}{l}{A = 4}\\{C = - 3}\end{array Show Step 5

To get the remaining constant we can use any value of $$x$$ and plug that along with the values of $$A$$ and $$C$$ 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 $$x = - \frac{3}{2}$$ or $$x = 2$$ since we used those 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*}7 & = \left( 4 \right){\left( { - 2} \right)^2} + B\left( 3 \right)\left( { - 2} \right) + \left( { - 3} \right)\left( 3 \right)\\ 7 & = 7 - 6B\\ 0 & = - 6B\hspace{0.25in} \Rightarrow \hspace{0.25in}B = 0\end{align*}

In this case one of the constants ended up being zero. This happens on occasion but there is no way, in general, to know ahead of time that was going to happen so don’t worry about it. If it turns out one of the constants is zero then we’ll figure that out when we do the work.

Show Step 6

The partial fraction decomposition is then,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{4{x^2} - 22x + 7}}{{\left( {2x + 3} \right){{\left( {x - 2} \right)}^2}}} = \frac{4}{{2x + 3}} - \frac{3}{{{{\left( {x - 2} \right)}^2}}}}}$