# Linear Factors

### Learning Outcomes

- Decompose a rational expression with distinct linear factors.
- Decompose a rational expression with repeated linear factors.

### recall how to add rational expressions

- Factor the numerator and denominator.
- Find the LCD of the expressions.
- Multiply the expressions by a form of 1 that changes the denominators to the LCD.
- Add or subtract the numerators over the common denominator.
- Simplify.

**partial fraction decomposition**, which is the undoing of the procedure to add or subtract rational expressions. In other words, it is a return from the single simplified

**rational expression**to the original expressions, called the

**partial fractions**. Some types of rational expressions require solving a system of equations in order to decompose them. For example, suppose we add the following fractions:

[latex]\dfrac{2}{x - 3}+\dfrac{-1}{x+2}[/latex]

We would first need to find a common denominator,[latex]\left(x+2\right)\left(x - 3\right)[/latex].

Next, we would write each expression with this common denominator and find the sum of the terms.[latex]\begin{align}&\frac{2}{x - 3}\left(\frac{x+2}{x+2}\right)+\frac{-1}{x+2}\left(\frac{x - 3}{x - 3}\right) \\[2mm] &=\frac{2x+4-x+3}{\left(x+2\right)\left(x - 3\right)} \\[2mm] &=\frac{x+7}{{x}^{2}-x - 6} \end{align}[/latex]

Partial fraction**decomposition**is the reverse of this procedure. We would start with the solution and rewrite (decompose) it as the sum of two fractions.

[latex]\begin{align} \dfrac{x+7}{{x}^{2}-x - 6}&=\dfrac{2}{x - 3}+\dfrac{-1}{x+2} \\[2mm]\text{Simplified sum}&\hspace{6mm}\text{Partial fraction decomposition} \end{align}[/latex]

We will investigate rational expressions with linear factors and quadratic factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator. When the denominator of the simplified expression contains distinct linear factors, it is likely that each of the original rational expressions, which were added or subtracted, had one of the linear factors as the denominator. In other words, using the example above, the factors of [latex]{x}^{2}-x - 6[/latex] are [latex]\left(x - 3\right)\left(x+2\right)[/latex], the denominators of the decomposed rational expression. So we will rewrite the simplified form as the sum of individual fractions and use a variable for each numerator. Then, we will solve for each numerator using one of several methods available for partial fraction decomposition.### tip for success

The techniques to decompose fractions are challenging and will require some practice to become familiar to you. The explanations of how to perform the decomposition may be confusing the first time (or more) that you read through them. If so, work through the given examples on paper first, slowly, step by step. Then return to the General Note and How To boxes to gain an understanding. It will take time and effort, so don't be discouraged if it takes multiple attempts for each example.### A General Note: Partial Fraction Decomposition of [latex]\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right)[/latex] Has Nonrepeated Linear Factors

The**partial fraction decomposition**of [latex]\dfrac{P\left(x\right)}{Q\left(x\right)}[/latex] when [latex]Q\left(x\right)[/latex] has nonrepeated linear factors 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)}{Q\left(x\right)}=\dfrac{{A}_{1}}{\left({a}_{1}x+{b}_{1}\right)}+\dfrac{{A}_{2}}{\left({a}_{2}x+{b}_{2}\right)}+\dfrac{{A}_{3}}{\left({a}_{3}x+{b}_{3}\right)}+\cdot \cdot \cdot +\dfrac{{A}_{n}}{\left({a}_{n}x+{b}_{n}\right)}[/latex].

### How To: Given a rational expression with distinct linear factors in the denominator, decompose it.

- Use a variable for the original numerators, usually [latex]A,B,[/latex] or [latex]C[/latex], depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use [latex]{A}_{n}[/latex] for each numerator
[latex]\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{{A}_{1}}{\left({a}_{1}x+{b}_{1}\right)}+\dfrac{{A}_{2}}{\left({a}_{2}x+{b}_{2}\right)}+\cdots \text{+}\dfrac{{A}_{n}}{\left({a}_{n}x+{b}_{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.

### recall how to eliminate denominators from an equation using the lcd

The following example relies upon the technique of multiplying both sides of a rational equation by the lowest common denominator (LCD) to eliminate the denominators from the equation.- Factor all denominators in the equation
- Find the LCD
- Multiply the whole equation by the LCD (each term). As you do, cancel out the denominator in each term. If the LCD has been chose correctly, there will be no denominators remaining at the end of this step.

### Example: Decomposing a Rational Expression with Distinct Linear Factors

Decompose the given**rational expression**with distinct linear factors.

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

Answer: We will separate the denominator factors and give each numerator a symbolic label, like [latex]A,B[/latex], or [latex]C[/latex].

[latex]\dfrac{3x}{\left(x+2\right)\left(x - 1\right)}=\dfrac{A}{\left(x+2\right)}+\dfrac{B}{\left(x - 1\right)}[/latex]

Multiply both sides of the equation by the common denominator to eliminate the fractions:[latex]\cancel{\left(x+2\right)}\cancel{\left(x - 1\right)}\left[\dfrac{3x}{\cancel{\left(x+2\right)}\cancel{\left(x - 1\right)}}\right]=\cancel{\left(x+2\right)}\left(x - 1\right)\left[\dfrac{A}{\cancel{\left(x+2\right)}}\right]+\left(x+2\right)\cancel{\left(x - 1\right)}\left[\dfrac{B}{\cancel{\left(x - 1\right)}}\right][/latex]

The resulting equation is[latex]3x=A\left(x - 1\right)+B\left(x+2\right)[/latex]

Expand the right side of the equation and collect like terms.[latex]\begin{gathered}3x=Ax-A+Bx+2B\\ 3x=\left(A+B\right)x-A+2B\end{gathered}[/latex]

Set up a system of equations associating corresponding coefficients.[latex]\begin{gathered}3=A+B\\ 0=-A+2B\end{gathered}[/latex]

### tip for success

Associating corresponding coefficients is a well-used mathematical technique. In this example, if[latex](3)x + 0 = (A+B)x + (-A+2B)[/latex]

then it must be true that[latex]3=A+B[/latex] and [latex]0=-A+2B[/latex].

[latex]\begin{align}3&=A+B \\ 0&=-A+2B \\ \hline 3&=0+3B \\[4mm] B&=1 \end{align}[/latex]

Substitute [latex]B=1[/latex] into one of the original equations in the system.[latex]\begin{align}3&=A+1\\ 2&=A\end{align}[/latex]

Thus, the partial fraction decomposition is[latex]\dfrac{3x}{\left(x+2\right)\left(x - 1\right)}=\dfrac{2}{\left(x+2\right)}+\dfrac{1}{\left(x - 1\right)}[/latex]

Another method to use to solve for [latex]A[/latex] or [latex]B[/latex] is by considering the equation that resulted from eliminating the fractions and substituting a value for [latex]x[/latex] that will make either the [latex]A-[/latex] or [latex]B-[/latex]term equal 0. If we let [latex]x=1[/latex], the [latex]A-[/latex] term becomes 0 and we can simply solve for [latex]B[/latex].[latex]\begin{align}3x&=A\left(x - 1\right)+B\left(x+2\right) \\ 3\left(1\right)&=A\left[\left(1\right)-1\right]+B\left[\left(1\right)+2\right] \\ 3&=0+3B\hfill \\ B&=1 \end{align}[/latex]

Next, either substitute [latex]B=1[/latex] into the equation and solve for [latex]A[/latex], or make the [latex]B-[/latex]term 0 by substituting [latex]x=-2[/latex] into the equation.[latex]\begin{align}3x&=A\left(x - 1\right)+B\left(x+2\right) \\ 3\left(-2\right)&=A\left[\left(-2\right)-1\right]+B\left[\left(-2\right)+2\right] \\ -6&=-3A+0 \\ \frac{-6}{-3}&=A \\ A&=2 \end{align}[/latex]

We obtain the same values for [latex]A[/latex] and [latex]B[/latex] using either method, so the decompositions are the same using either method.[latex]\dfrac{3x}{\left(x+2\right)\left(x - 1\right)}=\dfrac{2}{\left(x+2\right)}+\dfrac{1}{\left(x - 1\right)}[/latex]

Although this method is not seen very often in textbooks, we present it here as an alternative that may make some partial fraction decompositions easier. It is known as the**Heaviside method**, named after Charles Heaviside, a pioneer in the study of electronics.

### Try It

Find the partial fraction decomposition of the following expression.[latex]\dfrac{x}{\left(x - 3\right)\left(x - 2\right)}[/latex]

Answer: [latex-display]\dfrac{3}{x - 3}-\dfrac{2}{x - 2}[/latex-display]

[embed]## Decomposing P(x)/ Q(x), Where Q(x) Has Repeated Linear Factors

Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.### A General Note: Partial Fraction Decomposition of [latex]\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right)[/latex] Has Repeated Linear Factors

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 linear factor occurring [latex]n[/latex] times 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)}{Q\left(x\right)}=\dfrac{{A}_{1}}{\left(ax+b\right)}+\dfrac{{A}_{2}}{{\left(ax+b\right)}^{2}}+\dfrac{{A}_{3}}{{\left(ax+b\right)}^{3}}+\cdot \cdot \cdot +\dfrac{{A}_{n}}{{\left(ax+b\right)}^{n}}[/latex]

Write the denominator powers in increasing order.### How To: Given a rational expression with repeated linear factors, decompose it.

- Use a variable like [latex]A,B[/latex], or [latex]C[/latex] for the numerators and account for increasing powers of the denominators.
[latex]\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{{A}_{1}}{\left(ax+b\right)}+\dfrac{{A}_{2}}{{\left(ax+b\right)}^{2}}+ \text{. }\text{. }\text{. + }\dfrac{{A}_{n}}{{\left(ax+b\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 with Repeated Linear Factors

Decompose the given rational expression with repeated linear factors.[latex]\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}[/latex]

Answer: The denominator factors are [latex]x{\left(x - 2\right)}^{2}[/latex]. To allow for the repeated factor of [latex]\left(x - 2\right)[/latex], the decomposition will include three denominators: [latex]x,\left(x - 2\right)[/latex], and [latex]{\left(x - 2\right)}^{2}[/latex]. Thus,

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

Next, we multiply both sides by the common denominator.[latex]\begin{gathered}x{\left(x - 2\right)}^{2}\left[\dfrac{-{x}^{2}+2x+4}{x{\left(x - 2\right)}^{2}}\right]=\left[\dfrac{A}{x}+\dfrac{B}{\left(x - 2\right)}+\dfrac{C}{{\left(x - 2\right)}^{2}}\right]x{\left(x - 2\right)}^{2} \\[2mm] -{x}^{2}+2x+4=A{\left(x - 2\right)}^{2}+Bx\left(x - 2\right)+Cx \end{gathered}[/latex]

On the right side of the equation, we expand and collect like terms.[latex]-{x}^{2}+2x+4=A\left({x}^{2}-4x+4\right)+B\left({x}^{2}-2x\right)+Cx[/latex] [latex]\begin{align}&=A{x}^{2}-4Ax+4A+B{x}^{2}-2Bx+Cx \\ &=\left(A+B\right){x}^{2}+\left(-4A - 2B+C\right)x+4A \end{align}[/latex]

Next, we compare the coefficients of both sides. This will give the system of equations in three variables:[latex]-{x}^{2}+2x+4=\left(A+B\right){x}^{2}+\left(-4A - 2B+C\right)x+4A[/latex]

[latex]\begin{array}{rr}\hfill A+B=-1& \hfill \text{(1)}\\ \hfill -4A - 2B+C=2& \hfill \text{(2)}\\ \hfill 4A=4& \hfill \text{(3)}\end{array}[/latex]

Solving for [latex]A[/latex] , we have[latex]\begin{align}4A&=4 \\ A&=1 \end{align}[/latex]

Substitute [latex]A=1[/latex] into equation (1).[latex]\begin{align}A+B=-1 \\ \left(1\right)+B=-1 \\ B=-2 \end{align}[/latex]

Then, to solve for [latex]C[/latex], substitute the values for [latex]A[/latex] and [latex]B[/latex] into equation (2).[latex]\begin{align}-4A - 2B+C=2\\ -4\left(1\right)-2\left(-2\right)+C=2\\ -4+4+C=2\\ C=2\end{align}[/latex]

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

### Try It

Find the partial fraction decomposition of the expression with repeated linear factors.[latex]\dfrac{6x - 11}{{\left(x - 1\right)}^{2}}[/latex]

Answer: [latex-display]\dfrac{6}{x - 1}-\dfrac{5}{{\left(x - 1\right)}^{2}}[/latex-display]

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**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]. - Ex 1: Partial Fraction Decomposition (Linear Factors) .
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