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Study Guides > College Algebra CoRequisite Course

Quadratic Factors

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.

  1. 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]
  2. Multiply both sides of the equation by the common denominator to eliminate fractions.
  3. Expand the right side of the equation and collect like terms.
  4. 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]

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]\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]

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]

Answer: [latex-display]\dfrac{3}{x - 1}+\dfrac{2x - 4}{{x}^{2}+1}[/latex-display]

[embed]
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. https://youtu.be/prtx4o1wbaQ

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.

  1. 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]
  2. Multiply both sides of the equation by the common denominator to eliminate fractions.
  3. Expand the right side of the equation and collect like terms.
  4. 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]

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]\dfrac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}=\dfrac{A}{x}+\dfrac{Bx+C}{\left({x}^{2}+1\right)}+\dfrac{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{align}&{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 \\ &=A{x}^{4}+2A{x}^{2}+A+B{x}^{4}+B{x}^{2}+C{x}^{3}+Cx+D{x}^{2}+Ex \end{align}[/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{align}A+B=1 \\ C=1 \\ \\ 2A+B+D=1 \\ C+E=-1 \\ A=1 \end{align}[/latex]

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

[latex]\begin{align}1+B=1 \\ B=0 \end{align}[/latex]

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

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

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

[latex]\begin{align} 1+E=-1\\ E=-2\end{align}[/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]\dfrac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}=\dfrac{1}{x}+\dfrac{1}{\left({x}^{2}+1\right)}-\dfrac{x+2}{{\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]

Answer: [latex-display]\dfrac{x - 2}{{x}^{2}-2x+3}+\dfrac{2x+1}{{\left({x}^{2}-2x+3\right)}^{2}}[/latex-display]

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. https://youtu.be/Dupeou-FDnI

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  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].
  • Question ID 18139. Authored by: Shahbazian,Roy, mb McClure,Caren. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
  • Ex 5: Partial Fraction Decomposition (Linear and Quadratic Factors) . Authored by: Sousa, James (Mathispower4u.com). License: CC BY: Attribution.
  • Ex 6: Partial Fraction Decomposition (Repeating Quadratic Factors) . Authored by: Sousa, James (Mathispower4u.com). License: CC BY: Attribution.

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