Partial Fraction Decomposition: Learn It 3

Giselle Miller

Partial Fraction Decomposition: Learn It 3

Nonrepeated Irreducible Quadratic Factor

So far we have performed partial fraction decomposition with expressions that have had linear factors in the denominator, and we applied numerators [latex]A,B[/latex], or [latex]C[/latex] representing constants.

Now we will look at an example where one of the factors in the denominator is a quadratic expression that does not factor. This is referred to as an irreducible quadratic factor. In cases like this, we use a linear numerator such as [latex]Ax+B,Cx+D[/latex], etc.

partial fraction decomposition: nonrepeated irreducible quadratic factor

The partial fraction decomposition of [latex]\dfrac{P\left(x\right)}{Q\left(x\right)}[/latex] such that [latex]Q\left(x\right)[/latex] has a nonrepeated irreducible quadratic factor and the degree of [latex]P\left(x\right)[/latex] is less than the degree of [latex]Q\left(x\right)[/latex] is written as

[latex]\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{{A}_{1}x+{B}_{1}}{\left({a}_{1}{x}^{2}+{b}_{1}x+{c}_{1}\right)}+\dfrac{{A}_{2}x+{B}_{2}}{\left({a}_{2}{x}^{2}+{b}_{2}x+{c}_{2}\right)}+\cdot \cdot \cdot +\dfrac{{A}_{n}x+{B}_{n}}{\left({a}_{n}{x}^{2}+{b}_{n}x+{c}_{n}\right)}[/latex]

The decomposition may contain more rational expressions if there are linear factors. Each linear factor will have a different constant numerator: [latex]A,B,C[/latex], and so on.

How To: Given a rational expression where the factors of the denominator are distinct, irreducible quadratic factors, decompose it.

Use variables such as [latex]A,B[/latex], or [latex]C[/latex] for the constant numerators over linear factors, and linear expressions such as [latex]{A}_{1}x+{B}_{1},{A}_{2}x+{B}_{2}[/latex], etc., for the numerators of each quadratic factor in the denominator.
[latex]\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{A}{ax+b}+\dfrac{{A}_{1}x+{B}_{1}}{\left({a}_{1}{x}^{2}+{b}_{1}x+{c}_{1}\right)}+\dfrac{{A}_{2}x+{B}_{2}}{\left({a}_{2}{x}^{2}+{b}_{2}x+{c}_{2}\right)}+\cdot \cdot \cdot +\dfrac{{A}_{n}x+{B}_{n}}{\left({a}_{n}{x}^{2}+{b}_{n}x+{c}_{n}\right)}[/latex]
Multiply both sides of the equation by the common denominator to eliminate fractions.
Expand the right side of the equation and collect like terms.
Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.

Find a partial fraction decomposition of the given expression.

[latex]\dfrac{8{x}^{2}+12x - 20}{\left(x+3\right)\left({x}^{2}+x+2\right)}[/latex]

We have one linear factor and one irreducible quadratic factor in the denominator, so one numerator will be a constant and the other numerator will be a linear expression. Thus,

[latex]\dfrac{8{x}^{2}+12x - 20}{\left(x+3\right)\left({x}^{2}+x+2\right)}=\dfrac{A}{\left(x+3\right)}+\dfrac{Bx+C}{\left({x}^{2}+x+2\right)}[/latex]

We follow the same steps as in previous problems. First, clear the fractions by multiplying both sides of the equation by the common denominator.

[latex]\begin{align}\left(x+3\right)\left({x}^{2}+x+2\right)\left[\frac{8{x}^{2}+12x - 20}{\left(x+3\right)\left({x}^{2}+x+2\right)}\right]&=\left[\frac{A}{\left(x+3\right)}+\frac{Bx+C}{\left({x}^{2}+x+2\right)}\right]\left(x+3\right)\left({x}^{2}+x+2\right) \\[2mm] 8{x}^{2}+12x - 20&=A\left({x}^{2}+x+2\right)+\left(Bx+C\right)\left(x+3\right) \end{align}[/latex]

Notice we could easily solve for [latex]A[/latex] by choosing a value for [latex]x[/latex] that will make the [latex]Bx+C[/latex] term equal 0. Let [latex]x=-3[/latex] and substitute it into the equation.

[latex]\begin{align}8{x}^{2}+12x - 20&=A\left({x}^{2}+x+2\right)+\left(Bx+C\right)\left(x+3\right) \\ 8{\left(-3\right)}^{2}+12\left(-3\right)-20&=A\left({\left(-3\right)}^{2}+\left(-3\right)+2\right)+\left(B\left(-3\right)+C\right)\left(\left(-3\right)+3\right) \\ 16&=8A \\ A&=2 \end{align}[/latex]

Now that we know the value of [latex]A[/latex], substitute it back into the equation. Then expand the right side and collect like terms.

[latex]\begin{align} &8{x}^{2}+12x - 20=2\left({x}^{2}+x+2\right)+\left(Bx+C\right)\left(x+3\right) \\ &8{x}^{2}+12x - 20=2{x}^{2}+2x+4+B{x}^{2}+3B+Cx+3C \\ &8{x}^{2}+12x - 20=\left(2+B\right){x}^{2}+\left(2+3B+C\right)x+\left(4+3C\right) \end{align}[/latex]

Setting the coefficients of terms on the right side equal to the coefficients of terms on the left side gives the system of equations.

[latex]\begin{align}2+B=8 && \text{(1)} \\ 2+3B+C=12 && \text{(2)} \\ 4+3C=-20 && \text{(3)} \end{align}[/latex]

Solve for [latex]B[/latex] using equation (1) and solve for [latex]C[/latex] using equation (3).

[latex]\begin{align}2+B=8 && \text{(1)} \\ B=6 \\ \\ 4+3C=-20 && \text{(3)} \\ 3C=-24 \\ C=-8 \end{align}[/latex]

Thus, the partial fraction decomposition of the expression is

[latex]\dfrac{8{x}^{2}+12x - 20}{\left(x+3\right)\left({x}^{2}+x+2\right)}=\dfrac{2}{\left(x+3\right)}+\dfrac{6x - 8}{\left({x}^{2}+x+2\right)}[/latex]

Could we have just set up a system of equations to solve for [latex]A, B, \text{ and } C[/latex]?

Yes, we could have solved it by setting up a system of equations without solving for [latex]A[/latex] first. The expansion on the right would be:

[latex]\begin{align} 8{x}^{2}+12x - 20&=A{x}^{2}+Ax+2A+B{x}^{2}+3B+Cx+3C \\ 8{x}^{2}+12x - 20&=\left(A+B\right){x}^{2}+\left(A+3B+C\right)x+\left(2A+3C\right) \end{align}[/latex]

So the system of equations would be:

[latex]\begin{align}A+B=8 \\ A+3B+C=12 \\ 2A+3C=-20 \end{align}[/latex]

If you solve this system of equation, you will also find that [latex]A = 2, B = 6, \text{ and } C = -8[/latex].

Sometimes, a combination of methods is helpful to obtain the decomposition. In the example above, you could have set the system up first, then used [latex]x=-3[/latex] to obtain the value for [latex]A[/latex], and it would be quick work to obtain [latex]B[/latex] and [latex]C[/latex] from the system.Remember that creativity is a key component of doing mathematics and that there is often more than one good way to reach a conclusion.