### Learning Outcomes

- Decompose a rational expression with non-repeated irreducible quadratic factors.
- Decompose a rational expression that has repeated irreducible quadratic factors.

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,Bx+C[/latex], etc.

### tip for success

As with decomposing fractions with linear factors in the previous section, the techniques to decompose fractions with quadratic factors are challenging and will require some practice to become familiar to you.

Do work through the given examples on paper first, slowly, step by step before trying to make sense of the General Note and How To boxes.

It will take time and effort, so don’t be discouraged if it takes multiple attempts for each example.

### A General Note: Decomposition of [latex]\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right)[/latex] Has a 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.

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

### Q & A

#### Could we have just set up a system of equations to solve Example 3?

*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]

### tip for success

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.

### Try It

Find the partial fraction decomposition of the expression with a nonrepeating irreducible quadratic factor.

[latex]\dfrac{5{x}^{2}-6x+7}{\left(x - 1\right)\left({x}^{2}+1\right)}[/latex]

https://ohm.lumenlearning.com/multiembedq.php?id=18139&theme=oea&iframe_resize_id=mom1

In the following video, you will see another example of how to find the partial fraction decomposition for a rational expression that has quadratic factors.

## Decomposing P(x) / Q(x), When Q(x) Has a Repeated Irreducible Quadratic Factor

### tip for success

Do work through the example below carefully and more than once. The problem may appear to be complicated, but it relies on the same techniques you’ve already practiced. Keep your work well-organized to avoid making mistakes.

Now that we can decompose a simplified **rational expression** with an irreducible **quadratic** factor, we will learn how to do partial fraction decomposition when the simplified rational expression has repeated irreducible quadratic factors. The decomposition will consist of partial fractions with linear numerators over each irreducible quadratic factor represented in increasing powers.

### A General Note: Decomposition of [latex]\frac{P\left(x\right)}{Q\left(x\right)}[/latex] When *Q(x)* Has a Repeated Irreducible Quadratic Factor

The partial fraction decomposition of [latex]\dfrac{P\left(x\right)}{Q\left(x\right)}[/latex], when [latex]Q\left(x\right)[/latex] has a repeated 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

[latex]\dfrac{P\left(x\right)}{{\left(a{x}^{2}+bx+c\right)}^{n}}=\dfrac{{A}_{1}x+{B}_{1}}{\left(a{x}^{2}+bx+c\right)}+\dfrac{{A}_{2}x+{B}_{2}}{{\left(a{x}^{2}+bx+c\right)}^{2}}+\dfrac{{A}_{3}x+{B}_{3}}{{\left(a{x}^{2}+bx+c\right)}^{3}}+\cdot \cdot \cdot +\dfrac{{A}_{n}x+{B}_{n}}{{\left(a{x}^{2}+bx+c\right)}^{n}}[/latex]

Write the denominators in increasing powers.

### How To: Given a rational expression that has a repeated irreducible factor, decompose it.

- Use variables like [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 written in increasing powers, such as
[latex]\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{A}{ax+b}+\dfrac{{A}_{1}x+{B}_{1}}{\left(a{x}^{2}+bx+c\right)}+\dfrac{{A}_{2}x+{B}_{2}}{{\left(a{x}^{2}+bx+c\right)}^{2}}+\cdots +\text{ }\dfrac{{A}_{n}+{B}_{n}}{{\left(a{x}^{2}+bx+c\right)}^{n}}[/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.

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

### Try It

Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor.

[latex]\dfrac{{x}^{3}-4{x}^{2}+9x - 5}{{\left({x}^{2}-2x+3\right)}^{2}}[/latex]

This video provides you with another worked example of how to find the partial fraction decomposition for a rational expression that has repeating quadratic factors.