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Section 5.5 : Partial Fractions

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

\[\frac{{4{x^3} + 16x + 7}}{{{{\left( {{x^2} + 4} \right)}^2}}}\]

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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,

\[\frac{{4{x^3} + 16x + 7}}{{{{\left( {{x^2} + 4} \right)}^2}}} = \frac{{Ax + B}}{{{x^2} + 4}} + \frac{{Cx + D}}{{{{\left( {{x^2} + 4} \right)}^2}}}\] Show Step 2

The LCD for this expression is \({\left( {{x^2} + 4} \right)^2}\). Adding the terms back up gives,

\[\frac{{4{x^3} + 16x + 7}}{{{{\left( {{x^2} + 4} \right)}^2}}} = \frac{{\left( {Ax + B} \right)\left( {{x^2} + 4} \right) + Cx + D}}{{{{\left( {{x^2} + 4} \right)}^2}}}\] Show Step 3

Setting the numerators equal gives,

\[4{x^3} + 16x + 7 = \left( {Ax + B} \right)\left( {{x^2} + 4} \right) + Cx + D\] Show Step 4

Because we have an unfactorable quadratic equation here the method we used in the first problems from this section won’t work. So, we will need to multiply everything out and collect like terms.

\[\begin{align*}4{x^3} + 16x + 7 & = A{x^3} + 4Ax + B{x^2} + 4B + Cx + D\\ & = A{x^3} + B{x^2} + \left( {4A + C} \right)x + 4B + D\end{align*}\] Show Step 5

We now need to set coefficients equal. Remember this just means setting the coefficient of the \({x^3}\) on both sides equal and similarly for the coefficients of the \({x^2}\), the \(x\) and the constants. Doing this gives,

\[\begin{align*}A & = 4\\ B & = 0\\ 4A + C & = 16\\ 4B + D & = 7\end{align*}\] Show Step 6

As mentioned in the notes this is a system of equations that we really haven’t talked about how to solve in general, but that is not a real problem. From the first two equations we can see that we must have \(A = 4\) and \(B = 0\).

Once we have this we only need to plug that into the last two equations to determine the values of \(C\) and \(D\). Here is that work,

\[\begin{array}{l}{4\left( 4 \right) + C = 16}\\{4\left( 0 \right) + D = 7}\end{array}\hspace{0.25in} \to \hspace{0.25in}\begin{array}{l}{C = 0}\\{D = 7}\end{array}\]

In this case two 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, or more, of the constants are zero then we’ll figure that out when we do the work.

Show Step 7

The partial fraction decomposition is then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{4{x^3} + 16x + 7}}{{{{\left( {{x^2} + 4} \right)}^2}}} = \frac{{4x}}{{{x^2} + 4}} + \frac{7}{{{{\left( {{x^2} + 4} \right)}^2}}}}}\]