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Guias de estudo > College Algebra

Linear Factors

Learning Objectives

  • Decompose a rational expression with distinct linear factors
  • Decompose a rational expression with repeated linear factors
Recall the algebra regarding adding and subtracting rational expressions. These operations depend on finding a common denominator so that we can write the sum or difference as a single, simplified rational expression. In this section, we will look at 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, in case you were wondering what partial fractions has to do with linear systems. For example, suppose we add the following fractions:

2x−3+−1x+2\frac{2}{x - 3}+\frac{-1}{x+2}

We would first need to find a common denominator,

(x+2)(x−3)\left(x+2\right)\left(x - 3\right).

Next, we would write each expression with this common denominator and find the sum of the terms.

2x−3(x+2x+2)+−1x+2(x−3x−3)= 2x+4−x+3(x+2)(x−3)=x+7x2−x−6\begin{array}{l}\frac{2}{x - 3}\left(\frac{x+2}{x+2}\right)+\frac{-1}{x+2}\left(\frac{x - 3}{x - 3}\right)=\hfill \\ \text{ }\frac{2x+4-x+3}{\left(x+2\right)\left(x - 3\right)}=\frac{x+7}{{x}^{2}-x - 6}\hfill \end{array}

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.

x+7x2−x−6Simplified sum=2x−3+−1x+2Partial fraction decomposition\underset{\begin{array}{l}\\ \text{Simplified sum}\end{array}}{\frac{x+7}{{x}^{2}-x - 6}}=\underset{\begin{array}{l}\\ \text{Partial fraction decomposition}\end{array}}{\frac{2}{x - 3}+\frac{-1}{x+2}}

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 x2−x−6{x}^{2}-x - 6 are (x−3)(x+2)\left(x - 3\right)\left(x+2\right), 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.

A General Note: Partial Fraction Decomposition of P(x)Q(x):Q(x)\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right) Has Nonrepeated Linear Factors

The partial fraction decomposition of P(x)Q(x)\frac{P\left(x\right)}{Q\left(x\right)} when Q(x)Q\left(x\right) has nonrepeated linear factors and the degree of P(x)P\left(x\right) is less than the degree of Q(x)Q\left(x\right) is

P(x)Q(x)=A1(a1x+b1)+A2(a2x+b2)+A3(a3x+b3)+⋅⋅⋅+An(anx+bn)\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left({a}_{1}x+{b}_{1}\right)}+\frac{{A}_{2}}{\left({a}_{2}x+{b}_{2}\right)}+\frac{{A}_{3}}{\left({a}_{3}x+{b}_{3}\right)}+\cdot \cdot \cdot +\frac{{A}_{n}}{\left({a}_{n}x+{b}_{n}\right)}.

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

  1. Use a variable for the original numerators, usually A,B,A,B, or CC, depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use An{A}_{n} for each numerator
    P(x)Q(x)=A1(a1x+b1)+A2(a2x+b2)+⋯+An(anx+bn)\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left({a}_{1}x+{b}_{1}\right)}+\frac{{A}_{2}}{\left({a}_{2}x+{b}_{2}\right)}+\cdots \text{+}\frac{{A}_{n}}{\left({a}_{n}x+{b}_{n}\right)}
  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 Expression with Distinct Linear Factors

Decompose the given rational expression with distinct linear factors.

3x(x+2)(x−1)\frac{3x}{\left(x+2\right)\left(x - 1\right)}

Answer: We will separate the denominator factors and give each numerator a symbolic label, like A,B,A,B\text{\hspace{0.17em},} or CC.

3x(x+2)(x−1)=A(x+2)+B(x−1)\frac{3x}{\left(x+2\right)\left(x - 1\right)}=\frac{A}{\left(x+2\right)}+\frac{B}{\left(x - 1\right)}

Multiply both sides of the equation by the common denominator to eliminate the fractions:

(x+2)(x−1)[3x(x+2)(x−1)]=)(x+2)‾(x−1)[A)(x+2)‾]+(x+2))(x−1)‾[B)(x−1)‾]\left(x+2\right)\left(x - 1\right)\left[\frac{3x}{\left(x+2\right)\left(x - 1\right)}\right]=\overline{)\left(x+2\right)}\left(x - 1\right)\left[\frac{A}{\overline{)\left(x+2\right)}}\right]+\left(x+2\right)\overline{)\left(x - 1\right)}\left[\frac{B}{\overline{)\left(x - 1\right)}}\right]

The resulting equation is

3x=A(x−1)+B(x+2)3x=A\left(x - 1\right)+B\left(x+2\right)

Expand the right side of the equation and collect like terms.

3x=Ax−A+Bx+2B3x=(A+B)x−A+2B\begin{array}{l}3x=Ax-A+Bx+2B\\ 3x=\left(A+B\right)x-A+2B\end{array}

Set up a system of equations associating corresponding coefficients.

3=A+B0=−A+2B\begin{array}{l}3=A+B\\ 0=-A+2B\end{array}

Add the two equations and solve for BB.

\begin{array}\text{ }3=A+B \\ 0=-A+2B \hfill& \\ \text{_____________} \\ 3=0+3B \hfill& \\ 1=B \hfill& \end{array}

Substitute B=1B=1 into one of the original equations in the system.

3=A+12=A\begin{array}{l}3=A+1\\ 2=A\end{array}

Thus, the partial fraction decomposition is

3x(x+2)(x−1)=2(x+2)+1(x−1)\frac{3x}{\left(x+2\right)\left(x - 1\right)}=\frac{2}{\left(x+2\right)}+\frac{1}{\left(x - 1\right)}

Another method to use to solve for AA or BB is by considering the equation that resulted from eliminating the fractions and substituting a value for xx that will make either the A- or B-term equal 0. If we let x=1x=1, the A−A- term becomes 0 and we can simply solve for BB.

 3x=A(x−1)+B(x+2)3(1)=A[(1)−1]+B[(1)+2] 3=0+3B 1=B\begin{array}{l}\text{ }3x=A\left(x - 1\right)+B\left(x+2\right)\hfill \\ 3\left(1\right)=A\left[\left(1\right)-1\right]+B\left[\left(1\right)+2\right]\hfill \\ \text{ }3=0+3B\hfill \\ \text{ }1=B\hfill \end{array}

Next, either substitute B=1B=1 into the equation and solve for AA, or make the B-term 0 by substituting x=−2x=-2 into the equation.

 3x=A(x−1)+B(x+2) 3(−2)=A[(−2)−1]+B[(−2)+2] −6=−3A+0 −6−3=A 2=A\begin{array}{l}\text{ }3x=A\left(x - 1\right)+B\left(x+2\right)\hfill \\ \text{ }3\left(-2\right)=A\left[\left(-2\right)-1\right]+B\left[\left(-2\right)+2\right]\hfill \\ \text{ }-6=-3A+0\hfill \\ \text{ }\frac{-6}{-3}=A\hfill \\ \text{ 2}=A\hfill \end{array}

We obtain the same values for AA and BB using either method, so the decompositions are the same using either method.

3x(x+2)(x−1)=2(x+2)+1(x−1)\frac{3x}{\left(x+2\right)\left(x - 1\right)}=\frac{2}{\left(x+2\right)}+\frac{1}{\left(x - 1\right)}

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.

x(x−3)(x−2)\frac{x}{\left(x - 3\right)\left(x - 2\right)}

Answer: 3x−3−2x−2\frac{3}{x - 3}-\frac{2}{x - 2}

In this video, you will see another example of how to find a partial fraction decomposition when you have linear factors. https://youtu.be/WoVdOcuSI0I

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 P(x)Q(x):Q(x)\frac{P\left(x\right)}{Q\left(x\right)}:Q\left(x\right) Has Repeated Linear Factors

The partial fraction decomposition of P(x)Q(x)\frac{P\left(x\right)}{Q\left(x\right)}, when Q(x)Q\left(x\right) has a repeated linear factor occurring nn times and the degree of P(x)P\left(x\right) is less than the degree of Q(x)Q\left(x\right), is

P(x)Q(x)=A1(ax+b)+A2(ax+b)2+A3(ax+b)3+⋅⋅⋅+An(ax+b)n\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left(ax+b\right)}+\frac{{A}_{2}}{{\left(ax+b\right)}^{2}}+\frac{{A}_{3}}{{\left(ax+b\right)}^{3}}+\cdot \cdot \cdot +\frac{{A}_{n}}{{\left(ax+b\right)}^{n}}

Write the denominator powers in increasing order.

How To: Given a rational expression with repeated linear factors, decompose it.

  1. Use a variable like A,BA,B, or CC for the numerators and account for increasing powers of the denominators.
    P(x)Q(x)=A1(ax+b)+A2(ax+b)2+. . . + An(ax+b)n\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left(ax+b\right)}+\frac{{A}_{2}}{{\left(ax+b\right)}^{2}}+ \text{. }\text{. }\text{. + }\frac{{A}_{n}}{{\left(ax+b\right)}^{n}}
  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 with Repeated Linear Factors

Decompose the given rational expression with repeated linear factors.

−x2+2x+4x3−4x2+4x\frac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}

Answer: The denominator factors are x(x−2)2x{\left(x - 2\right)}^{2}. To allow for the repeated factor of (x−2)\left(x - 2\right), the decomposition will include three denominators: x,(x−2)x,\left(x - 2\right), and (x−2)2{\left(x - 2\right)}^{2}. Thus, −x2+2x+4x3−4x2+4x=Ax+B(x−2)+C(x−2)2\frac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}=\frac{A}{x}+\frac{B}{\left(x - 2\right)}+\frac{C}{{\left(x - 2\right)}^{2}} Next, we multiply both sides by the common denominator.

x(x−2)2[−x2+2x+4x(x−2)2]=[Ax+B(x−2)+C(x−2)2]x(x−2)2 −x2+2x+4=A(x−2)2+Bx(x−2)+Cx\begin{array}{l}x{\left(x - 2\right)}^{2}\left[\frac{-{x}^{2}+2x+4}{x{\left(x - 2\right)}^{2}}\right]=\left[\frac{A}{x}+\frac{B}{\left(x - 2\right)}+\frac{C}{{\left(x - 2\right)}^{2}}\right]x{\left(x - 2\right)}^{2}\hfill \\ \text{ }-{x}^{2}+2x+4=A{\left(x - 2\right)}^{2}+Bx\left(x - 2\right)+Cx\hfill \end{array}

On the right side of the equation, we expand and collect like terms.

−x2+2x+4=A(x2−4x+4)+B(x2−2x)+Cx =Ax2−4Ax+4A+Bx2−2Bx+Cx =(A+B)x2+(−4A−2B+C)x+4A\begin{array}{l}-{x}^{2}+2x+4=A\left({x}^{2}-4x+4\right)+B\left({x}^{2}-2x\right)+Cx\hfill \\ \text{ }=A{x}^{2}-4Ax+4A+B{x}^{2}-2Bx+Cx\hfill \\ \text{ }=\left(A+B\right){x}^{2}+\left(-4A - 2B+C\right)x+4A\hfill \end{array}

Next, we compare the coefficients of both sides. This will give the system of equations in three variables:

−x2+2x+4=(A+B)x2+(−4A−2B+C)x+4A-{x}^{2}+2x+4=\left(A+B\right){x}^{2}+\left(-4A - 2B+C\right)x+4A

A+B=−1(1)−4A−2B+C=2(2)4A=4(3)\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}

Solving for AA , we have

4A=4 A=1\begin{array}{l}4A=4\hfill \\ \text{ }A=1\hfill \end{array}

Substitute A=1A=1 into equation (1).

 A+B=−1(1)+B=−1 B=−2\begin{array}{l}\text{ }A+B=-1\hfill \\ \left(1\right)+B=-1\hfill \\ \text{ }B=-2\hfill \end{array}

Then, to solve for CC, substitute the values for AA and BB into equation (2).

 −4A−2B+C=2−4(1)−2(−2)+C=2 −4+4+C=2 C=2\begin{array}{r}\hfill \text{ }-4A - 2B+C=2\\ \hfill -4\left(1\right)-2\left(-2\right)+C=2\\ \hfill \text{ }-4+4+C=2\\ \hfill \text{ }C=2\end{array}

Thus,

−x2+2x+4x3−4x2+4x=1x−2(x−2)+2(x−2)2\frac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}=\frac{1}{x}-\frac{2}{\left(x - 2\right)}+\frac{2}{{\left(x - 2\right)}^{2}}

Try It

Find the partial fraction decomposition of the expression with repeated linear factors.

6x−11(x−1)2\frac{6x - 11}{{\left(x - 1\right)}^{2}}

Answer: 6x−1−5(x−1)2\frac{6}{x - 1}-\frac{5}{{\left(x - 1\right)}^{2}}

In this video, you will see an example of how to find the partial fraction decomposition of a rational expression with repeated linear factors. https://youtu.be/6DdwGw_5dvk

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