Example: Decomposing [latex]\frac{P\left(x\right)}{Q\left(x\right)}[/latex] When Q(x) Contains a Nonrepeated Irreducible Quadratic Factor

Find a partial fraction decomposition of the given expression.

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

Answer: 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]\frac{8{x}^{2}+12x - 20}{\left(x+3\right)\left({x}^{2}+x+2\right)}=\frac{A}{\left(x+3\right)}+\frac{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{array}{l}\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)\hfill \\ \text{ }8{x}^{2}+12x - 20=A\left({x}^{2}+x+2\right)+\left(Bx+C\right)\left(x+3\right)\hfill \end{array}[/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{array}{l}\text{ }8{x}^{2}+12x - 20=A\left({x}^{2}+x+2\right)+\left(Bx+C\right)\left(x+3\right)\hfill \\ \text{ }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)\hfill \\ \text{ }16=8A\hfill \\ \text{ }A=2\hfill \end{array}[/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{array}{l}\hfill \\ 8{x}^{2}+12x - 20=2\left({x}^{2}+x+2\right)+\left(Bx+C\right)\left(x+3\right)\hfill \\ 8{x}^{2}+12x - 20=2{x}^{2}+2x+4+B{x}^{2}+3B+Cx+3C\hfill \\ 8{x}^{2}+12x - 20=\left(2+B\right){x}^{2}+\left(2+3B+C\right)x+\left(4+3C\right)\hfill \end{array}[/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{array}{ll}\text{ }2+B=8\hfill & \text{(1)}\hfill \\ 2+3B+C=12\hfill & \text{(2)}\hfill \\ \text{ }4+3C=-20\hfill & \text{(3)}\hfill \end{array}[/latex]

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

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

Thus, the partial fraction decomposition of the expression is

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

Example: Decomposing a Rational Function with a Repeated Irreducible Quadratic Factor in the Denominator

Decompose the given expression that has a repeated irreducible factor in the denominator.

[latex]\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}[/latex]

Answer: The factors of the denominator are [latex]x,\left({x}^{2}+1\right)[/latex], and [latex]{\left({x}^{2}+1\right)}^{2}[/latex]. Recall that, when a factor in the denominator is a quadratic that includes at least two terms, the numerator must be of the linear form [latex]Ax+B[/latex]. So, let’s begin the decomposition.

[latex]\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}=\frac{A}{x}+\frac{Bx+C}{\left({x}^{2}+1\right)}+\frac{Dx+E}{{\left({x}^{2}+1\right)}^{2}}[/latex]

We eliminate the denominators by multiplying each term by [latex]x{\left({x}^{2}+1\right)}^{2}[/latex]. Thus,

[latex]{x}^{4}+{x}^{3}+{x}^{2}-x+1=A{\left({x}^{2}+1\right)}^{2}+\left(Bx+C\right)\left(x\right)\left({x}^{2}+1\right)+\left(Dx+E\right)\left(x\right)[/latex]

Expand the right side.

[latex]\begin{array}{l} \text{ }{x}^{4}+{x}^{3}+{x}^{2}-x+1=A\left({x}^{4}+2{x}^{2}+1\right)+B{x}^{4}+B{x}^{2}+C{x}^{3}+Cx+D{x}^{2}+Ex\hfill \\ \text{ }=A{x}^{4}+2A{x}^{2}+A+B{x}^{4}+B{x}^{2}+C{x}^{3}+Cx+D{x}^{2}+Ex\hfill \end{array}[/latex]

Now we will collect like terms.

[latex]{x}^{4}+{x}^{3}+{x}^{2}-x+1=\left(A+B\right){x}^{4}+\left(C\right){x}^{3}+\left(2A+B+D\right){x}^{2}+\left(C+E\right)x+A[/latex]

Set up the system of equations matching corresponding coefficients on each side of the equal sign.

[latex]\begin{array}{l}\text{ }A+B=1\hfill \\ \text{ }C=1\hfill \\ 2A+B+D=1\hfill \\ \text{ }C+E=-1\hfill \\ \text{ }A=1\hfill \end{array}[/latex]

We can use substitution from this point. Substitute [latex]A=1[/latex] into the first equation.

[latex]\begin{array}{l}1+B=1\hfill \\ \text{ }B=0\hfill \end{array}[/latex]

Substitute [latex]A=1[/latex] and [latex]B=0[/latex] into the third equation.

[latex]\begin{array}{l}2\left(1\right)+0+D=1\hfill \\ \text{ }D=-1\hfill \end{array}[/latex]

Substitute [latex]C=1[/latex] into the fourth equation.

[latex]\begin{array}{r}\hfill 1+E=-1\\ \hfill \text{ }E=-2\end{array}[/latex]

Now we have solved for all of the unknowns on the right side of the equal sign. We have [latex]A=1[/latex], [latex]B=0[/latex], [latex]C=1[/latex], [latex]D=-1[/latex], and [latex]E=-2[/latex]. We can write the decomposition as follows:

[latex]\frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}=\frac{1}{x}+\frac{1}{\left({x}^{2}+1\right)}-\frac{x+2}{{\left({x}^{2}+1\right)}^{2}}[/latex]