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Section 5.3 Partial-Fraction Decomposition

Georg Simon Ohm (1787–1854)

Georg Ohm was born in Erlangen, Germany. His father, Johann, a mechanic and a self-educated man, was interested in philosophy and mathematics and gave his children an excellent education in physics, chemistry, mathematics, and philosophy through his own teachings. The Ohm’s law named in Georg Ohm’s honor appeared in his famous pamphlet Die galvanische Kette, mathematisch bearbeitet (1827). Ohm’s work was not appreciated in Germany, where it was first published. Eventually, it was recognized, first by the British Royal Society, which awarded Ohm the Copley Medal in 1841. In 1849, Ohm became a curator of the Bavarian Academy’s science museum and began to lecture at the University of Munich. Finally, in 1852 (two years before his death), Ohm achieved his lifelong ambition of becoming chair of physics at the University of Munich.

Before Starting this Section, Review

1 Division of polynomials (Section 3.3 , page 353)
2 Factoring polynomials (Section P.4 )
3 Irreducible polynomials (Section P.4 , page 42)
4 Rational expressions (Section P.5 , page 50)

Objectives

1 Become familiar with partial-fraction decomposition.
2 Decompose P(x)Q(x) $\frac{P (x)}{Q (x)}$ when Q(x) has only distinct linear factors.
3 Decompose $\frac{P (x)}{Q (x)}$ when Q(x) has repeated linear factors.
4 Decompose $\frac{P (x)}{Q (x)}$ when Q(x) has distinct irreducible quadratic factors.
5 Decompose $\frac{P (x)}{Q (x)}$ when Q(x) has repeated irreducible quadratic factors.

Ohm’s Law

If a voltage is applied across a resistor (such as a wire), a current will flow. Ohm’s law states that in a circuit at constant temperature,

$I = \frac{V}{R}$

where V is the voltage (measured in volts), I is the current (measured in amperes), and R is the resistance (measured in ohms). A parallel circuit consists of two or more resistors connected as shown in Figure 5.9. In the figure, the current I from the source divides into two currents, $I_{1}$ and $I_{2},$ with $I_{1}$ going through one resistor and $I_{2}$ going through the other, so that $I = I_{1} + I_{2} .$ Suppose we want to find the total resistance R in the circuit due to the two resistors $R_{1}$ and $R_{2}$ connected in parallel. Ohm’s law can be used to show that in the parallel circuit in Figure 5.9, $\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} .$ This result is called the parallel property of resistors. In Example 8, we examine the parallel property of resistors by using partial fractions.

Partial Fractions

1 Become familiar with partial-fraction decomposition.

Recall that to add two rational expressions such as $\frac{2}{x + 3}$ and $\frac{3}{x - 1}$ required you to find the least common denominator, rewrite expressions with the same denominator, and then add the corresponding numerators. This procedure gives

$\frac{2}{x + 3} + \frac{3}{x - 1} = \frac{2 (x - 1)}{(x + 3) (x - 1)} + \frac{3 (x + 3)}{(x - 1) (x + 3)} = \frac{5 x + 7}{(x + 3) (x - 1)} .$

In some applications of algebra and more advanced mathematics, you need a reverse procedure for splitting a fraction such as $\frac{5 x + 7}{(x - 1) (x + 3)}$ into the simpler fractions to obtain:

$\frac{5 x + 7}{(x + 3) (x - 1)} = \frac{2}{x + 3} + \frac{3}{x - 1} .$

Each of the two fractions on the right is called a partial fraction. Their sum is called the partial-fraction decomposition of the rational expression on the left.

A rational expression $\frac{P (x)}{Q (x)}$ is called improper if the degree $P (x) \geq degree Q (x)$ and is called proper if the degree $P (x) < degree Q (x) .$ Examples of improper rational expressions are

$\frac{x^{3}}{x^{2} - 1}, \frac{(x + 2) (x - 1)}{(x - 2) (x + 3)}, and \frac{x^{2} + x + 1}{2 x + 3} .$

Using long division, we can express an improper rational fraction $\frac{P (x)}{Q (x)}$ as the sum of a polynomial and a proper rational fraction:

$\begin{array}{c} \frac{P (x)}{Q (x)} & = & S (x) & + & \frac{R (x)}{Q (x)} . \\ ↑ & ↑ & ↑ \\ \begin{array}{l} Improper \\ Fraction \end{array} & \begin{array}{l} Polynomial \end{array} & \begin{array}{l} Proper \\ Fraction \end{array} \end{array}$

We therefore restrict our discussion of the decomposition of $\frac{P (x)}{Q (x)}$ into partial fractions to cases involving proper rational fractions. We also assume that P(x) and Q(x) have no common factor.

The problem of decomposing a proper fraction $\frac{P (x)}{Q (x)}$ into partial fractions depends on the type of factors in the denominator Q(x). In this section, we consider four cases for Q(x).

`Q`(`x`) Has Only Distinct Linear Factors

2 Decompose $\frac{P (x)}{Q (x)}$ when Q(x) has only distinct linear factors.

We can find the constants $A_{1}, A_{2}, \dots, A_{n}$ by using the following procedure. Note that if the number of constants is small, we use the letters $A, B, C, \dots,$ instead of $A_{1}, A_{2}, A_{3}, \dots .$

Procedure in Action

Example 1 Partial-Fraction Decomposition

OBJECTIVE	EXAMPLE
Find the partial-fraction decomposition of a rational expression. Step 1 Write the form of the partial-fraction decomposition with the unknown constants $A, B, C, \dots$ in the numerators of the decomposition.	Find the partial-fraction decomposition of $\frac{3 x + 26}{(x - 3) (x + 4)} .$ $\frac{3 x + 26}{(x - 3) (x + 4)} = \frac{A}{x - 3} + \frac{B}{x + 4}$
Step 2 Multiply both sides of the equation in Step 1 by the original denominator. Use the distributive property and eliminate common factors. Simplify.	Multiply both sides by $(x - 3) (x + 4)$ and simplify to obtain $3 x + 26 = (x + 4) A + (x - 3) B .$
Step 3 Write both sides of the equation in Step 2 in descending powers of `x` and equate the coefficients of like powers of `x`.	$3 x + 26 = (A + B) x + (4 A - 3 B)$ $\begin{array}{rcl} {\begin{cases} 3 = A + B \\ 26 = 4 A - 3 B \end{cases} & \begin{array}{l} Equate coefficients of x . \\ Equate constant coefficients . \end{array} \end{array}$
Step 4 Solve the linear system resulting from Step 3 for the constants $A, B, C, \dots$	Solving the system of equations in Step 3, we obtain $A = 5$ and $B = - 2 .$
Step 5 Substitute the values you found for $A, B, C, \dots$ into the equation in Step 1 and write the partial-fraction decomposition.	$\begin{array}{rcl} \frac{3 x + 26}{(x - 3) (x + 4)} & = & \frac{5}{x - 3} + \frac{- 2}{x + 4} & \begin{array}{l} Replace A with 5 and \\ B with - 2 in Step 1. \end{array} \\ \frac{3 x + 26}{(x - 3) (x + 4)} & = & \frac{5}{x - 3} - \frac{2}{x + 4} \end{array}$

Practice Problem 1

Find the partial-fraction decomposition of $\frac{2 x - 7}{(x + 1) (x - 2)} .$

Example 2 Finding the Partial-Fraction Decomposition When the Denominator Has Only Distinct Linear Factors

Find the partial-fraction decomposition of the expression

$\frac{10 x - 4}{x^{3} - 4 x} .$

Solution

First, factor the denominator.

$\begin{array}{rcl} x^{3} - 4 x & = & x (x^{2} - 4) & Distributive property \\ = & x (x - 2) (x + 2) & x^{2} - 4 = (x - 2) (x + 2) \end{array}$

Each of the factors $x, x - 2,$ and $x + 2$ and becomes a denominator for a partial fraction.

Step 1 The partial-fraction decomposition is given by

$\begin{array}{rcl} \frac{10 x - 4}{x (x - 2) (x + 2)} & = & \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 2} & \begin{array}{l} Use A, B, and C instead of \\ A_{1}, A_{2}, and A_{3} . \end{array} \end{array}$
Step 2 Multiply both sides by the common denominator $x (x - 2) (x + 2) .$
Step 3 Now use the fact that two equal polynomials have equal corresponding coefficients. Writing $10 x - 4 = 0 x^{2} + 10 x - 4,$ we have

$0 x^{2} + 10 x - 4 = (A + B + C) x^{2} + (2 B - 2 C) x - 4 A .$

Equating corresponding coefficients leads to the system of equations.

$\begin{array}{rcl} {\begin{array}{rcl} A & + & B & + & C & = & 0 \\ 2 B & - & 2 C & = & 10 \\ - 4 A & = & - 4 \end{array} & \begin{array}{l} Equate coefficients of x^{2} . \\ Equate coefficients of x . \\ Equate constant coefficients . \end{array} \end{array}$
Step 4 Solve the system of equations in Step 3 to obtain $A = 1, B = 2,$ and $C = - 3 .$ (See Exercise 75 .)
Step 5 The partial-fraction decomposition is

$\frac{10 x - 4}{x^{3} - 4 x} = \frac{1}{x} + \frac{2}{x - 2} + \frac{- 3}{x + 2} = \frac{1}{x} + \frac{2}{x - 2} - \frac{3}{x + 2} .$

Alternative Solution of Example 2

In Example 2, to find the constants A, B, C, we equated coefficients of like powers of x and solved the resulting system. An alternative (and sometimes quicker) method is to substitute well-chosen values for x in equation (1) found in Step 2:

$10 x - 4 = A (x - 2) (x + 2) + B x (x + 2) + C x (x - 2) (1)$

Substitute $x = 2$ in equation (1) to cause the terms containing A and C to be 0.

$\begin{array}{rcl} 10 (2) - 4 & = & A (2 - 2) (2 + 2) + B (2) (2 + 2) + C (2) (2 - 2) \\ 16 & = & 8 B & Simplify . \\ 2 & = & B & Solve for B . \end{array}$

Now substitute $x = - 2$ in equation (1) to get

$\begin{array}{rcl} 10 (- 2) - 4 & = & A (- 2 - 2) (- 2 + 2) + B (- 2) (- 2 + 2) + C (- 2) (- 2 - 2) \\ - 24 & = & 8 C & Simplify . \\ - 3 & = & C & Solve for C . \end{array}$

Finally, substitute $x = 0$ in equation (1) to get

$\begin{array}{rcl} 10 (0) - 4 & = & A (0 - 2) (0 + 2) + B (0) (0 + 2) + C (0) (0 - 2) \\ - 4 & = & - 4 A & Simplify . \\ 1 & = & A & Solve for A . \end{array}$

Because we found the same values for A, B, and C, the partial-fraction decomposition will also be the same.

Practice Problem 2

Find the partial-fraction decomposition of

$\frac{3 x^{2} + 4 x + 3}{x^{3} - x} .$

`Q`(`x`) Has Repeated Linear Factors

3 Decompose $\frac{P (x)}{Q (x)}$ when Q(x) has repeated linear factors.

Example 3 Finding the Partial-Fraction Decomposition When the Denominator Has Repeated Linear Factors

Find the partial-fraction decomposition of $\frac{x + 4}{(x + 3) {(x - 1)}^{2}}$ .

Solution

Step 1 The linear factor $(x - 1)$ is repeated twice, and the factor $(x + 3)$ is nonrepeating. So the partial-fraction decomposition has the form:

$\frac{x + 4}{(x + 3) {(x - 1)}^{2}} = \frac{A}{x + 3} + \frac{B}{x - 1} + \frac{C}{{(x - 1)}^{2}} .$
Steps 2–4 Multiply both sides of the equation in Step 1 by the original denominator, $(x + 3) {(x - 1)}^{2};$ then use the distributive property on the right side and simplify to get the identity:

$x + 4 = A {(x - 1)}^{2} + B (x + 3) (x - 1) + C (x + 3) (1)$

Substitute $x = 1$ in equation (1) to obtain:

$1 + 4 = A {(1 - 1)}^{2} + B (1 + 3) (1 - 1) + C (1 + 3) .$

This simplifies to $5 = 4 C,$ so $C = \frac{5}{4} .$

To find A, we substitute $x = - 3$ in equation (1) to get

$- 3 + 4 = A {(- 3 - 1)}^{2} + B (- 3 + 3) (- 3 - 1) + C (- 3 + 3) .$

This simplifies to $1 = 16 A,$ so $A = \frac{1}{16} .$

To find B, we replace x with any convenient number, say, 0 in equation (1). We have

$\begin{array}{rcl} 0 + 4 & = & A {(0 - 1)}^{2} + B (0 + 3) (0 - 1) + C (0 + 3) \\ 4 & = & A - 3 B + 3 C (2) & Simplify . \end{array}$

Now replace A with $\frac{1}{16}$ and C with $\frac{5}{4}$ in equation (2) to get:

$\begin{array}{rcl} 4 & = & \frac{1}{16} - 3 B + 3 (\frac{5}{4}) \\ 64 & = & 1 - 48 B + 60 & Multiply both sides by 16 . \\ B & = & - \frac{1}{16} & Solve for B . \end{array}$

We have $A = \frac{1}{16}, B = - \frac{1}{16},$ and $C = \frac{5}{4} .$
Step 5 Substitute $A = \frac{1}{16}, B = - \frac{1}{16},$ and $C = \frac{5}{4}$ in the decomposition in Step 1 to get

$\frac{x + 4}{(x + 3) {(x - 1)}^{2}} = \frac{\frac{1}{16}}{x + 3} + \frac{- \frac{1}{16}}{x - 1} + \frac{\frac{5}{4}}{{(x - 1)}^{2}}$

or

$\frac{x + 4}{(x + 3) {(x - 1)}^{2}} = \frac{1}{16 (x + 3)} - \frac{1}{16 (x - 1)} + \frac{5}{4 {(x - 1)}^{2}}$

Practice Problem 3

Find the partial-fraction decomposition of

$\frac{x + 5}{x {(x - 1)}^{2}} .$

Example 4 Finding the Partial-Fraction Decomposition When the Denominator Has Repeated Linear Factors

Find the partial-fraction decomposition of $\frac{2 x^{2} - 7 x + 9}{{(x - 1)}^{3}} .$

Solution

Step 1 The only linear factor of the denominator, $x - 1,$ is repeated 3 times. The partial-fraction decomposition has the form:

$\frac{2 x^{2} - 7 x + 9}{{(x - 1)}^{3}} = \frac{A}{x - 1} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{3}} (1)$
Step 2 Multiply both sides of equation (1) by ${(x - 1)}^{3}$ and simplify to obtain the identity:

$2 x^{2} - 7 x + 9 = A {(x - 1)}^{2} + B (x - 1) + C (2)$ .
Step 3 Substitute $x = 1$ in equation (2) to get $C = 4 .$

So, equation (2) becomes:

$\begin{array}{rcl} 2 x^{2} - 7 x + 9 & = & A {(x - 1)}^{2} + B (x - 1) + 4 \\ 2 x^{2} - 7 x + 5 & = & A {(x - 1)}^{2} + B (x - 1) (3) & Subtract 4 from both sides . \end{array}$
Step 4 We still need to find A and B. We can substitute any number for x in equation (3). We choose two numbers (other than 1) to obtain two equations with unknowns A and B.

$\begin{array}{rcl} \begin{array}{l} Letting x = 0 leads to \\ Letting x = - 1, after simplifying, leads to \end{array} & {\begin{array}{l} 5 & = & A - B & (4) . \\ 7 & = & 2 A - B & (5) . \end{array} \end{array}$

We solve equations (4) and (5) to get $A = 2$ and $B = - 3$ .
Step 5 We have $A = 2, B = - 3,$ and $C = 4$ . Substituting these values in equation (1), we have the partial-fraction decomposition

$\frac{2 x^{2} - 7 x + 9}{{(x - 1)}^{3}} = \frac{2}{x - 1} - \frac{3}{{(x - 1)}^{2}} + \frac{4}{{(x - 1)}^{3}} .$

Practice Problem 4

Find the partial-fraction decomposition of $\frac{2 x + 1}{{(x + 3)}^{2}} .$

`Q`(`x`) Has Distinct Irreducible Quadratic Factors

4 Decompose $\frac{P (x)}{Q (x)}$ when Q(x) has distinct irreducible quadratic factors.

Example 5 Finding the Partial-Fraction Decomposition When the Denominator Has a Distinct Irreducible Quadratic Factor

Find the partial-fraction decomposition of $\frac{3 x^{2} - 8 x + 1}{(x - 4) (x^{2} + 1)}$ .

Solution

Step 1 The factor $x - 4$ is linear and $x^{2} + 1$ is irreducible; so the partial-fraction decomposition has the form:

$\frac{3 x^{2} - 8 x + 1}{(x - 4) (x^{2} + 1)} = \frac{A}{x - 4} + \frac{B x + C}{x^{2} + 1} .$
Step 2 Multiply both sides of the decomposition in Step 1 by the original denominator, $(x - 4) (x^{2} + 1),$ and simplify to get

$3 x^{2} - 8 x + 1 = A (x^{2} + 1) + (B x + C) (x - 4) .$

Substitute $x = 4$ to find $17 = 17 A,$ or $A = 1 .$
Step 3 Collect like terms and write both sides of the equation in Step 2 in descending powers of x.

$3 x^{2} - 8 x + 1 = (A + B) x^{2} + (- 4 B + C) x + (A - 4 C)$

Equate corresponding coefficients to get

$\begin{array}{rcl} {\begin{array}{rcl} A & + & B & = & 3 & (1) \\ - 4 B & + & C & = & - 8 & (2) \\ A & - & 4 C & = & 1 & (3) \end{array} & \begin{array}{l} Equate the coefficients of x^{2} . \\ Equate the coefficients of x . \\ Equate the constant coefficients . \end{array} \end{array}$
Step 4 Substitute $A = 1$ from Step 2 in equation (1) to get $1 + B = 3,$ or $B = 2 .$

Substitute $A = 1$ in equation (3) to get $1 - 4 C = 1,$ or $C = 0 .$
Step 5 Substitute $A = 1, B = 2,$ and $C = 0$ into the decomposition in Step 1 to get:

$\frac{3 x^{2} - 8 x + 1}{(x - 4) (x^{2} + 1)} = \frac{1}{x - 4} + \frac{2 x}{x^{2} + 1} .$

Practice Problem 5

Find the partial-fraction decomposition of

$\frac{3 x^{2} + 5 x - 2}{x (x^{2} + 2)} .$

`Q`(`x`) Has Repeated Irreducible Quadratic Factors

5 Decompose $\frac{P (x)}{Q (x)}$ when Q(x) has repeated irreducible quadratic factors.

Example 6 Finding the Partial-Fraction Decomposition When the Denominator Has a Repeated Irreducible Quadratic Factor

Find the partial-fraction decomposition of $\frac{2 x^{4} - x^{3} + 13 x^{2} - 2 x + 13}{(x - 1) {(x^{2} + 4)}^{2}} .$

Solution

Step 1 The denominator has a nonrepeating linear factor $(x - 1)$ and an irreducible quadratic factor $(x^{2} + 4)$ that appears twice. The decomposition therefore has the form:

$\frac{2 x^{4} - x^{3} + 13 x^{2} - 2 x + 13}{(x - 1) {(x^{2} + 4)}^{2}} = \frac{A}{x - 1} + \frac{B x + C}{x^{2} + 4} + \frac{D x + E}{{(x^{2} + 4)}^{2}} .$
Step 2 Multiply both sides of the decomposition in Step 1 by the original denominator $(x - 1) {(x^{2} + 4)}^{2},$ use the distributive property, and eliminate common factors:

$\begin{array}{l} 2 x^{4} - x^{3} + 13 x^{2} - 2 x + 13 = \\ A {(x^{2} + 4)}^{2} + (B x + C) (x - 1) (x^{2} + 4) + (D x + E) (x - 1) . \end{array}$

Substitute $x = 1$ and simplify to obtain $25 = 25 A,$ or $A = 1 .$
Steps 3–4 Multiply the factors on the right side of the equation in Step 2 and collect like terms to obtain

$\begin{array}{l} 2 x^{4} - x^{3} + 13 x^{2} - 2 x + 13 = \\ (A + B) x^{4} + (- B + C) x^{3} + (8 A + 4 B - C + D) x^{2} + (- 4 B + 4 C - D + E) x + (16 A - 4 C - E) . \end{array}$

Equate corresponding coefficients to get the system:

$\begin{array}{rcl} {\begin{array}{rcl} A & + & B & = & 2 & (1) \\ - & B & + & C & = & - 1 & (2) \\ 8 A & + & 4 B & - & C & + & D & = & 13 & (3) \\ - & 4 B & + & 4 C & - & D & + & E & = & - 2 & (4) \\ 16 A & - & 4 C & - & E & = & 13 & (5) \end{array} & \begin{array}{l} Coefficient of x^{4} \\ Coefficient of x^{3} \\ Coefficient of x^{2} \\ Coefficient of x \\ Constant coefficient \end{array} \end{array}$

Now back-substitute $A = 1$ from Step 2 in equation (1) to get $B = 1 .$

Again, back-substitute $B = 1$ in equation (2) to get $C = 0 .$

Back-substitute $A = 1$ and $C = 0$ in equation (5) to obtain $E = 3 .$ Again, back-substitute $A = 1, B = 1,$ and $C = 0$ in equation (3) to get $D = 1 .$ Hence, $A = 1, B = 1, C = 0, D = 1,$ and $E = 3 .$
Step 5 Substitute $A = 1, B = 1, C = 0, D = 1,$ and $E = 3$ in the equation in Step 1 to obtain the partial-fraction decomposition.

$\frac{2 x^{4} - x^{3} + 13 x^{2} - 2 x + 13}{(x - 1) {(x^{2} + 4)}^{2}} = \frac{1}{x - 1} + \frac{x}{x^{2} + 4} + \frac{x + 3}{{(x^{2} + 4)}^{2}}$

Practice Problem 6

Find the partial-fraction decomposition of

$\frac{x^{2} + 3 x + 1}{{(x^{2} + 1)}^{2}} .$

Example 7 Learning a Clever Way to Add

Express the sum

$\frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \dots + \frac{1}{2019 \cdot 2020}$

as a fraction of whole numbers in lowest terms.

Solution

Each term in the sum is of the form: $\frac{1}{k (k + 1)} .$ From the partial-fraction decomposition, we have:

$\frac{1}{k (k + 1)} = \frac{1}{k} - \frac{1}{k + 1} .$

Therefore, by using this decomposition for each term, we get:

$\begin{array}{l} \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \dots + \frac{1}{2019 \cdot 2020} \\ = (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + \dots + (\frac{1}{2019} - \frac{1}{2020}) . \end{array}$

After removing parentheses, we see that most terms cancel in pairs $(such as - \frac{1}{3} and \frac{1}{3}),$ leaving only the first and the last term. The original sum is therefore referred to as a telescoping sum that “collapses” to

$\frac{1}{2} - \frac{1}{2020} = \frac{1010 - 1}{2020} = \frac{1009}{2020} .$

Practice Problem 7

Express the sum $\frac{1}{4 \cdot 5} + \frac{1}{5 \cdot 6} + \frac{1}{6 \cdot 7} + \dots + \frac{1}{3111 \cdot 3112}$ as a fraction of whole numbers in lowest terms.

Example 8 Calculating the Resistance in a Circuit

The total resistance R due to two resistors connected in parallel is given by

$R = \frac{x (x + 5)}{3 x + 10} .$

Use partial-fraction decomposition to write $\frac{1}{R}$ in terms of partial fractions. What does each term represent?

Solution

From $R = \frac{x (x + 5)}{3 x + 10},$ we get the following:

$\begin{array}{rcl} \frac{1}{R} & = & \frac{3 x + 10}{x (x + 5)} & Take the reciprocal of both sides . \\ \frac{3 x + 10}{x (x + 5)} & = & \frac{A}{x} + \frac{B}{x + 5} & Form of partial-fraction decomposition \\ 3 x + 10 & = & A (x + 5) + B x & Multiply both sides by x (x + 5) and simplify . \end{array}$

Substitute $x = 0$ in the last equation to obtain $5 A = 10,$ or $A = 2 .$ Now substitute $x = - 5$ in the same equation to get $- 5 = - 5 B,$ or $B = 1 .$ Thus, we have

$\begin{array}{rcl} \frac{1}{R} & = & \frac{3 x + 10}{x (x + 5)} = \frac{2}{x} + \frac{1}{x + 5} \\ \frac{1}{R} & = & \frac{1}{\frac{x}{2}} + \frac{1}{x + 5} Rewrite the right side . \end{array}$

The last equation says that if two resistors with resistances $R_{1} = \frac{x}{2}$ and $R_{2} = x + 5$ are connected in parallel, they will produce a total resistance R, where $R = \frac{x (x + 5)}{3 x + 10} .$

Practice Problem 8

Repeat Example 8, assuming that $R = \frac{(x + 1) (x + 2)}{4 x + 7} .$

Section 5.3 Exercises

Concepts and Vocabulary

In a rational expression, if the degree of the numerator, P(x), is less than the degree of the denominator, Q(x), then the expression is a fraction.
The numerators in the partial-fraction decomposition of a rational expression are constants if the denominator, Q(x), can be factored into .
In a rational expression, if $(x - 8)$ is a linear factor that is repeated three times in the denominator, then the portion of the expression’s partial-fraction decomposition that corresponds to ${(x - 8)}^{3}$ has terms.
True or False. The form of the partial-fraction decomposition of a rational expression depends only on its denominator.
True or False. Both ${(x - 1)}^{2}$ and ${(x + 2)}^{2}$ are irreducible quadratic factors of the polynomial $Q (x) = {(x - 1)}^{2} {(x + 2)}^{2} .$
True or False. If the denominator of a rational expression is $Q (x) = x^{3} - 9 x$ then x, $x - 3,$ and $x + 3$ become denominators in its partial-fraction decomposition.
True or False. If $(x - 2)$ is a factor that is repeated exactly 3 times in Q(x), then $(x - 2), {(x - 2)}^{2}$ , and ${(x - 2)}^{3}$ are denominators in the partial-fraction decomposition of $\frac{P (x)}{Q (x)} .$
True or False. The quadratic expression $x^{2} + 2 x + 1$ is irreducible.

Building Skills

In Exercises 9–18, write the form of the partial-fraction decomposition of each rational expression. You do not need to solve for the constants.

$\frac{1}{(x - 1) (x + 2)}$
$\frac{x}{(x + 1) (x - 3)}$
$\frac{1}{x^{2} + 7 x + 6}$
$\frac{3}{x^{2} - 6 x + 8}$
$\frac{2}{x^{3} - x^{2}}$
$\frac{x - 1}{{(x + 2)}^{2} (x - 3)}$
$\frac{x^{2} - 3 x + 3}{(x + 1) (x^{2} - x + 1)}$
$\frac{2 x + 3}{{(x - 1)}^{2} (x^{2} + x + 1)}$
$\frac{3 x - 4}{{(x^{2} + 1)}^{2}}$
$\frac{x - 2}{{(2 x + 3)}^{2} {(x^{2} + 3)}^{2}}$

In Exercises 19–66, find the partial-fraction decomposition of each rational expression.

$\frac{2 x + 1}{(x + 1) (x + 2)}$
$\frac{7}{(x - 2) (x + 5)}$
$\frac{1}{x^{2} + 4 x + 3}$
$\frac{x}{x^{2} + 5 x + 6}$
$\frac{2}{x^{2} + 2 x}$
$\frac{x + 9}{x^{2} - 9}$
$\frac{x}{(x + 1) (x + 2) (x + 3)}$
$\frac{8 x - 14}{(x - 1) (x + 2) (x - 3)}$
$\frac{x^{2}}{(x - 1) (x^{2} + 5 x + 4)}$
$\frac{2 x + 14}{(x + 1) (x^{2} - 6 x + 5)}$
$\frac{3 x + 1}{(x - 5) {(x + 3)}^{2}}$
$\frac{1 - 2 x}{(x + 3) {(x - 4)}^{2}}$
$\frac{8 x + 1}{{(x + 2)}^{2} (x - 3)}$
$\frac{3 x + 2}{x^{2} (x + 1)}$
$\frac{x - 1}{{(x + 1)}^{2}}$
$\frac{x}{{(x - 2)}^{2}}$
$\frac{x - 1}{{(2 x - 3)}^{2}}$
$\frac{6 x + 1}{{(3 x + 1)}^{2}}$
$\frac{2 x^{2} + x}{{(x + 1)}^{3}}$
$\frac{x^{2} + 2}{{(x - 1)}^{3}}$
$\frac{2 x + 3}{{(x + 2)}^{3}}$
$\frac{x^{2} - 8 x + 18}{{(x - 5)}^{3}}$
$\frac{- x^{2} + 3 x + 1}{x^{3} + 2 x^{2} + x}$
$\frac{5 x^{2} - 8 x + 2}{x^{3} - 2 x^{2} + x}$
$\frac{3 x + 1}{(x + 1) (x^{2} - 1)}$
$\frac{6 x - 4}{(x - 2) (x^{2} - 4)}$
$\frac{1}{x^{2} {(x + 1)}^{2}}$
$\frac{2}{{(x - 1)}^{2} {(x + 3)}^{2}}$
$\frac{1}{{(x^{2} - 1)}^{2}}$
$\frac{3}{{(x^{2} - 4)}^{2}}$
$\frac{x^{2} - 5}{(x - 2) (x^{2} - 2 x + 3)}$
$\frac{x^{2} + 2 x + 1}{(x + 2) (x^{2} + x + 2)}$
$\frac{x - 1}{{(x + 1)}^{2} (x^{2} + 2 x + 2)}$
$\frac{x^{2} - 3}{{(x - 2)}^{2} (x^{2} - x + 1)}$
$\frac{x^{2} + 1}{(x - 2) (x^{2} - 3 x + 2)}$
$\frac{x + 1}{x (2 x^{2} - 5 x - 3)}$
$\frac{6 x + 7}{4 x^{2} + 12 x + 9}$
$\frac{2 x + 3}{9 x^{2} + 30 x + 25}$
$\frac{x - 3}{x^{3} + x^{2}}$
$\frac{x^{2} + 1}{x^{4} - x^{3}}$
$\frac{x^{2} + 2 x + 4}{x^{3} + x^{2}}$
$\frac{x^{2} + 2 x - 1}{(x - 1) (x^{2} + 1)}$
$\frac{x}{x^{4} - 1}$
$\frac{3 x^{3} - 5 x^{2} + 12 x + 4}{x^{4} - 16}$
$\frac{1}{x {(x^{2} + 1)}^{2}}$
$\frac{x^{2}}{{(x^{2} + 2)}^{2}}$
$\frac{2 x^{2} + 3 x}{(x^{2} + 1) (x^{2} + 2)}$
$\frac{x^{2} - 2 x}{(x^{2} + 9) (x^{2} + x + 7)}$

Applying the Concepts

In Exercises 67–70, express each sum as a fraction of whole numbers in lowest terms.

$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n (n + 1)}$
$\frac{2}{1 \cdot 3} + \frac{2}{2 \cdot 4} + \frac{2}{3 \cdot 5} + \dots + \frac{2}{100 \cdot 102}$
$\frac{2}{1 \cdot 3} + \frac{2}{3 \cdot 5} + \frac{2}{5 \cdot 7} + \dots + \frac{2}{(2 n - 1) (2 n + 1)}$
$\frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \dots + \frac{2}{100 \cdot 101 \cdot 102}$

$\begin{array}{l} [Hint : Write \\ \begin{array}{rcl} \frac{2}{k (k + 1) (k + 2)} & = & \frac{1}{k} - \frac{2}{k + 1} + \frac{1}{k + 2} \\ = & \frac{1}{k} - \frac{1}{k + 1} - \frac{1}{k + 1} + \frac{1}{k + 2} .] \end{array} \end{array}$

In Exercises 71–74, the total resistance R in a circuit is given. Use partial-fraction decomposition to write $\frac{1}{R}$ in terms of partial fractions. Interpret each term in the partial fraction.

Circuit resistance. $R = \frac{(x + 1) (x + 3)}{2 x + 4}$
Circuit resistance. $R = \frac{(x + 2) (x + 4)}{7 x + 20}$
Circuit resistance. $R = \frac{R_{1} R_{2} R_{3}}{R_{1} R_{2} + R_{2} R_{3} + R_{3} R_{1}}$
Circuit resistance. $R = \frac{x (x + 2) (x + 4)}{3 x^{2} + 12 x + 8}$

Beyond the Basics

Solve the system of equations in Example 2 to verify the values of the constants A, B, and C.

In Exercises 76–81, find the partial-fraction decomposition of each rational expression.

$\frac{1}{x^{3} + 1}$
$\frac{4 x}{{(x^{2} - 1)}^{2}}$
$\frac{2 x + 3}{(x^{2} + 1) (x + 1)}$
$\frac{x + 1}{(x^{2} + 1) {(x - 1)}^{2}}$
$\frac{x}{{(x^{2} + 1)}^{2} (x - 1)}$
$\frac{x^{3}}{{(x + 1)}^{2} {(x + 2)}^{2}}$
Find the partial-fraction decomposition of

$\frac{4 x}{x^{4} + 4} .$
$\begin{array}{l} [Hint : x^{4} + 4 = x^{4} + 4 + 4 x^{2} - 4 x^{2} = {(x^{2} + 2)}^{2} - {(2 x)}^{2} \\ = (x^{2} - 2 x + 2) (x^{2} + 2 x + 2) .] \end{array}$

In Exercises 83 and 84, find the partial-fraction decomposition by first using long division.

$\frac{x^{2} + 4 x + 5}{x^{2} + 3 x + 2}$
$\frac{(x - 1) (x + 2)}{(x + 3) (x - 4)}$
Show that $\begin{array}{l} \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{9}} + \dots \\ + \frac{1}{\sqrt{2 n + 1} + \sqrt{2 n + 3}} = \frac{\sqrt{2 n + 3} - \sqrt{3}}{2} . \end{array}$
Show that $\begin{array}{l} \frac{1}{1 \cdot 3 \cdot 5} + \frac{1}{3 \cdot 5 \cdot 7} + \frac{1}{5 \cdot 7 \cdot 9} + \dots \\ + \frac{1}{(2 n - 1) (2 n + 1) (2 n + 3)} = \frac{n (n + 2)}{3 (2 n + 1) (2 n + 3)} . \end{array}$

Critical Thinking / Discussion / Writing

In the partial-fraction decomposition of an expression, we obtain a system of equations by equating coefficients of the powers of x. Justify the validity of this step.
Let $f (x) = \frac{x + 8}{x^{2} + x - 2} = g (x) + h (x),$ where g(x) and h(x) are the terms in the partial-fraction decomposition of f(x).
1. Sketch the graphs of f, g, and h.
2. Compare the graphs of f and g. What do you observe?
3. Compare the graphs of f and h. What do you observe?

Getting Ready for the Next Section

In Exercises 89–92, solve each equation for x.

$x^{2} + 10 x + 21 = 0$
$x^{2} - 2 x - 24 = 0$
$2 x^{2} + x - 10 = 0$
$6 x^{2} + x - 2 = 0$

In Exercises 93–96, solve each system of equations by the substitution method.

${\begin{array}{rcl} x + y & = & 3 \\ 3 x + 2 y & = & 7 \end{array}$
${\begin{array}{rcl} 3 x + y & = & - 3 \\ 4 x + 3 y & = & 1 \end{array}$
${\begin{array}{rcl} x - 2 y & = & 1 \\ 2 x + 3 y & = & 16 \end{array}$
${\begin{array}{rcl} x - 2 y & = & 10 \\ \frac{1}{2} x + \frac{1}{3} y & = & 1 \end{array}$

In Exercises 97–100, solve each system of equations by the elimination method.

${\begin{array}{rcl} 2 x - y & = & 4 \\ 3 x + 2 y & = & 13 \end{array}$
${\begin{array}{rcl} 2 x + 3 y & = & - 1 \\ 3 x - 2 y & = & 5 \end{array}$
${\begin{array}{rcl} 2 x + 5 y & = & 1 \\ 3 x - 2 y & = & - 8 \end{array}$
${\begin{array}{rcl} 2 x - 3 y & = & 6 \\ - 3 x + 2 y & = & 1 \end{array}$

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Table of Contents for Section 5.3 Partial-Fraction Decomposition

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Table of Contents for
Section 5.3 Partial-Fraction Decomposition