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### Section 5.5 : Partial Fractions

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

$\frac{{125 + 4x - 9{x^2}}}{{\left( {x - 1} \right)\left( {x + 3} \right)\left( {x + 4} \right)}}$

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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,

$\frac{{125 + 4x - 9{x^2}}}{{\left( {x - 1} \right)\left( {x + 3} \right)\left( {x + 4} \right)}} = \frac{A}{{x - 1}} + \frac{B}{{x + 3}} + \frac{C}{{x + 4}}$ Show Step 2

The LCD for this expression is $$\left( {x - 1} \right)\left( {x + 3} \right)\left( {x + 4} \right)$$. Adding the terms back up gives,

$\frac{{125 + 4x - 9{x^2}}}{{\left( {x - 1} \right)\left( {x + 3} \right)\left( {x + 4} \right)}} = \frac{{A\left( {x + 3} \right)\left( {x + 4} \right) + B\left( {x - 1} \right)\left( {x + 4} \right) + C\left( {x - 1} \right)\left( {x + 3} \right)}}{{\left( {x - 1} \right)\left( {x + 3} \right)\left( {x + 4} \right)}}$ Show Step 3

Setting the numerators equal gives,

$125 + 4x - 9{x^2} = A\left( {x + 3} \right)\left( {x + 4} \right) + B\left( {x - 1} \right)\left( {x + 4} \right) + C\left( {x - 1} \right)\left( {x + 3} \right)$ Show Step 4

Now all we need to do is pick “good” values of $$x$$ to determine the constants. Here is that work.

\begin{array}{l}{x = - 4:}\\{x = - 3:}\\{x = 1:}\end{array}\hspace{0.25in}\begin{aligned} - 35 & = 5C\\32 &= - 4B\\120 & = 20A\end{aligned}\hspace{0.25in} \to \hspace{0.25in}\begin{array}{l}{A = 6}\\{B = - 8}\\{C = - 7}\end{array} Show Step 5

The partial fraction decomposition is then,

$\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{125 + 4x - 9{x^2}}}{{\left( {x - 1} \right)\left( {x + 3} \right)\left( {x + 4} \right)}} = \frac{6}{{x - 1}} - \frac{8}{{x + 3}} - \frac{7}{{x + 4}}}}$