4-5 Partial Fractions
★
72. Average Cost. The total cost of producing x units of a certain product is given by
C(x) 15x2 2x 2,000
(C) Find the dimensions of the pen that will require the
least amount of fencing material.
(D) Graph the function L, including any asymptotes.
★
The average cost per unit for producing x units is
C(x) C(x)/x.
(A) Find the rational function C.
(B) At what production level will the average cost per unit
be minimal?
(C) Sketch the graph of C, including any asymptotes.
★
74. Construction. Rework Problem 73 with the added assumption that the pen is to be divided into two sections, as shown
in the figure.
x
x
73. Construction. A rectangular dog pen is to be made to enclose an area of 225 square feet.
(A) If x represents the width of the pen, express the total
length L(x) of the fencing material required for the pen
in terms of x.
(B) Considering the physical limitations, what is the domain of the function L?
SECTION
4-5
337
x
Partial Fractions
• Basic Theorems
• Partial Fraction Decomposition
You have now had considerable experience combining two or more rational expressions into a single rational expression. For example, problems such as
2
3
2(x 4) 3(x 5)
5x 7
x5 x4
(x 5)(x 4)
(x 5)(x 4)
should seem routine. Frequently in more advanced courses, particularly in calculus,
it is advantageous to be able to reverse this process—that is, to be able to express a
rational expression as the sum of two or more simpler rational expressions called partial fractions. As is often the case with reverse processes, the process of decomposing a rational expression into partial fractions is more difficult than combining rational expressions. Basic to the process is the factoring of polynomials, so the topics
discussed earlier in this chapter can be put to effective use.
We confine our attention to rational expressions of the form P(x)/D(x), where P(x)
and D(x) are polynomials with real coefficients and no common factors. In addition,
we assume that the degree of P(x) is less than the degree of D(x). If the degree of P(x)
is greater than or equal to that of D(x), we have only to divide P(x) by D(x) to obtain
R(x)
P(x)
Q(x) D(x)
D(x)
where the degree of R(x) is less than that of D(x). For example,
x4 3x3 2x2 5x 1
6x 2
x2 x 1 2
2
x 2x 1
x 2x 1
338
4 Polynomial and Rational Functions
If the degree of P(x) is less than that of D(x), then P(x)/D(x) is called a proper
fraction.
• Basic Theorems
Our task now is to establish a systematic way to decompose a proper fraction into the
sum of two or more partial fractions. The following three theorems take care of the
problem completely. Theorems 1 and 3 are stated without proof.
Theorem 1
Equal Polynomials
Two polynomials are equal to each other if and only if the coefficients of terms
of like degree are equal.
For example, if
Equate the constant terms.
(A 2B)x B 5x 3
aFEDDbDDEFc
Equate the coefficients of x.
then
B 3
Substitute B 3 into the second equation to solve for A.
A 2B 5
A 2(3) 5
A 11
EXPLORE-DISCUSS 1
If
x 5 A(x 1) B(x 3)
(1)
is a polynomial identity (that is, both sides represent the same polynomial), then
equating coefficients produces the system
1 A 0B
Equating coefficients of x
5 A 3B
Equating constant terms
(A) Solve this system (see Section 2-2).
(B) For an alternate method of solution, substitute x 3 in (1) to find A and then
substitute x 1 in (1) to find B. Explain why method B is valid.
4-5 Partial Fractions
Theorem 2
339
Linear and Quadratic Factor Theorem
For a polynomial with real coefficients, there always exists a complete factoring involving only linear and/or quadratic factors with real coefficients where
the linear and quadratic factors are prime relative to the real numbers.
That Theorem 2 is true can be seen as follows: From earlier theorems in this
chapter, we know that an nth-degree polynomial P(x) has n zeros and n linear factors. The real zeros of P(x) correspond to linear factors of the form (x r), where r
is a real number. Since P(x) has real coefficients, the imaginary zeros occur in conjugate pairs. Thus, the imaginary zeros correspond to pairs of factors of the form
[x (a bi)] and [x (a bi)], where a and b are real numbers. Multiplying these
two imaginary factors, we have
[x (a bi)][x (a bi)] x2 2ax a2 b2
This quadratic polynomial with real coefficients is a factor of P(x). Thus, P(x)
can be factored into a product of linear factors and quadratic factors, all with real
coefficients.
• Partial Fraction
Decomposition
Theorem 3
We are now ready to state Theorem 3, which forms the basis for partial fraction
decomposition.
Partial Fraction Decomposition
Any proper fraction P(x)/D(x) reduced to lowest terms can be decomposed into
the sum of partial fractions as follows:
1. If D(x) has a nonrepeating linear factor of the form ax b, then the partial
fraction decomposition of P(x)/D(x) contains a term of the form
A
ax b
A a constant
2. If D(x) has a k-repeating linear factor of the form (ax b)k, then the partial fraction decomposition of P(x)/D(x) contains k terms of the form
A2
A1
Ak
ax b (ax b)2
(ax b)k
A1, A2, . . . , Ak constants
3. If D(x) has a nonrepeating quadratic factor of the form ax2 bx c, which
is prime relative to the real numbers, then the partial fraction decomposition
of P(x)/D(x) contains a term of the form
Ax B
ax bx c
2
A, B constants
340
4 Polynomial and Rational Functions
4. If D(x) has a k-repeating quadratic factor of the form (ax2 bx c)k, where
ax2 bx c is prime relative to the real numbers, then the partial fraction
decomposition of P(x)/D(x) contains k terms of the form
A2x B2
Ak x Bk
A1x B1
ax2 bx c (ax2 bx c)2
(ax2 bx c)k
A1, . . . , Ak,
B1, . . . , Bk constants
Let’s see how the theorem is used to obtain partial fraction decompositions in
several examples.
EXAMPLE 1
Nonrepeating Linear Factors
Decompose into partial fractions:
Solution
5x 7
x 2x 3
2
We first try to factor the denominator. If it can’t be factored in the real numbers, then
we can’t go any further. In this example, the denominator factors, so we apply part 1
from Theorem 3:
A
B
5x 7
(x 1)(x 3) x 1 x 3
(2)
To find the constants A and B, we combine the fractions on the right side of equation
(2) to obtain
5x 7
A(x 3) B(x 1)
(x 1)(x 3)
(x 1)(x 3)
Since these fractions have the same denominator, their numerators must be equal.
Thus,
5x 7 A(x 3) B(x 1)
(3)
We could multiply the right side and find A and B by using Theorem 1, but in this
case it is easier to take advantage of the fact that equation (3) is an identity—that is,
it must hold for all values of x. In particular, we note that if we let x 1, then the
second term of the right side drops out and we can solve for A:
5 1 7 A(1 3) B(1 1)
12 4A
A3
Similarly, if we let x 3, the first term drops out and we find
4-5 Partial Fractions
341
8 4B
B2
Hence,
5x 7
3
2
x2 2x 3 x 1 x 3
(4)
as can easily be checked by adding the two fractions on the right.
Matched Problem 1
EXPLORE-DISCUSS 2
Decompose into partial fractions:
7x 6
x x6
2
A graphing utility can also be used to check a partial fraction decomposition. To
check Example 1, we graph the left and right sides of (4) in a graphing utility (Fig.
1). Discuss how the Trace feature on the graphing utility can be used to check that
the graphing utility is displaying two identical graphs.
10
FIGURE 1
10
10
10
EXAMPLE 2
Repeating Linear Factors
Decompose into partial fractions:
Solution
6x2 14x 27
(x 2)(x 3)2
Using parts 1 and 2 from Theorem 3, we write
B
C
6x2 14x 27
A
2
(x 2)(x 3)
x 2 x 3 (x 3)2
A(x 3)2 B(x 2)(x 3) C(x 2)
(x 2)(x 3)2
Thus, for all x,
6x2 14x 27 A(x 3)2 B(x 2)(x 3) C(x 2)
342
4 Polynomial and Rational Functions
If x 3, then
If x 2, then
15 5C
25 25A
C 3
A1
There are no other values of x that will cause terms on the right to drop out. Since
any value of x can be substituted to produce an equation relating A, B, and C, we let
x 0 and obtain
27 9A 6B 2C
Substitute A 1 and C 3.
27 9 6B 6
B5
Thus,
6x2 14x 27
1
5
3
2
(x 2)(x 3)
x 2 x 3 (x 3)2
Matched Problem 2
EXAMPLE 3
Decompose into partial fractions:
Nonrepeating Linear and Quadratic Factors
Decompose into partial fractions:
Solution
x2 11x 15
(x 1)(x 2)2
5x2 8x 5
(x 2)(x2 x 1)
First, we see that the quadratic in the denominator can’t be factored further in the real
numbers. Then, we use parts 1 and 3 from Theorem 3 to write
5x2 8x 5
A
Bx C
(x 2)(x2 x 1) x 2 x2 x 1
A(x2 x 1) (Bx C)(x 2)
(x 2)(x2 x 1)
Thus, for all x,
5x2 8x 5 A(x2 x 1) (Bx C)(x 2)
If x 2, then
9 3A
A3
4-5 Partial Fractions
343
If x 0, then, using A 3, we have
5 3 2C
C 1
If x 1, then, using A 3 and C 1, we have
2 3 (B 1)(1)
B2
Hence,
3
2x 1
5x2 8x 5
2
2
(x 2)(x x 1) x 2 x x 1
Matched Problem 3
Decompose into partial fractions:
EXAMPLE 4
Repeating Quadratic Factors
Decompose into partial fractions:
x3 4x2 9x 5
(x2 2x 3)2
Since x2 2x 3 can’t be factored further in the real numbers, we proceed to use
part 4 from Theorem 3 to write
x3 4x2 9x 5
Ax B
Cx D
2
2
2
2
(x 2x 3)
x 2x 3 (x 2x 3)2
(Ax B)(x2 2x 3) Cx D
(x2 2x 3)2
Thus, for all x,
x3 4x2 9x 5 (Ax B)(x2 2x 3) Cx D
Since the substitution of carefully chosen values of x doesn’t lead to the immediate
determination of A, B, C, or D, we multiply and rearrange the right side to obtain
x3 4x2 9x 5 Ax3 (B 2A)x2 (3A 2B C)x (3B D)
Now we use Theorem 1 to equate coefficients of terms of like degree:
3B D 5
abc
3A 2B C 9
1x3
abc
4x 2
9x
5
abc
abc
abc
aGEDbDEGc
B 2A 4
aGGFEbEFGGc
A1
aGEDbDEGc
Solution
7x2 11x 6
(x 1)(2x2 3x 2)
Ax3 (B 2A)x 2 (3A 2B C )x (3B D)
344
4 Polynomial and Rational Functions
From these equations we easily find that A 1, B 2, C 2, and D 1. Now
we can write
x3 4x2 9x 5
2x 1
x2
2
(x2 2x 3)2
x 2x 3 (x2 2x 3)2
Matched Problem 4
Decompose into partial fractions:
3x3 6x2 7x 2
(x2 2x 2)2
Answers to Matched Problems
4
3
3
2
1
2.
x2 x3
x 1 x 2 (x 2)2
3x
x2
4. 2
x 2x 2 (x2 2x 2)2
1.
EXERCISE
3.
2
3x 2
x 1 2x2 3x 2
4-5
A
B
In Problems 1–4, find constants A and B so that the indicated equation is a polynomial identity.
In Problems 11–22, decompose into partial fractions.
11.
3x 40
x2 x 12
12.
4x 43
x2 3x 10
13.
8x 22
6x2 17x 14
14.
x 27
10x2 13x 3
15.
6x2 14x 4
4x3 4x2 x
16.
7x 8
9x3 12x2 4x
17.
18.
5x2 7x 6
B
C
A
(x 1)(x 2)2 x 1 x 2 (x 2)2
10x2 4x 3
2x3 x2 x
9x2 14
x x2 2x
19.
20.
A
B
C
x2 5x 12
(x 1)(x 3)2 x 1 x 3 (x 3)2
4x3 5x2 6x 5
x4 2x2 1
2x3 x2 11x
x4 8x2 16
21.
4x2 5x 3
x3 x 2
22.
5x2 8x 12
x3 x2 4
1. 5x 10 A(x 4) B(x 6)
2. 3x 18 A(x 1) B(x 2)
3. 9x 5 A(2x 5) B(3x 2)
4. 32x 13 A(4x 7) B(2x 8)
In Problems 5–10, find constants A, B, C, and D so that the
right side is equal to the left.
5.
6.
7.
x2 5x 3 A Bx C
2
x(x2 1)
x
x 1
8.
3x2 2x 16 A Bx C
2
x(x2 4)
x
x 4
Ax B
2x3 x2 2
Cx D
9. 2
(x x 2)2 x2 x 2 (x2 x 2)2
4x 2x 13
Cx D
Ax B
2
2
2
2
(x x 4)
x x 4 (x x 4)2
2
10.
3
C
In Problems 23–26, decompose into a sum of a polynomial
and partial fractions.
23.
x5 4x2 16x 40
x4 2x3 8x 16
24.
2x5 2x4 6x3
x4 4x 3
25.
2x5 7x2 20x 21
x4 4x2 12x 9
26.
x5 2x4 3x2 14x
x4 x2 4x 4
4-5 Partial Fractions
345
In Problems 27–30, a and b are real constants. Decompose
into partial fractions.
27.
x
,a0
(x a)2
29.
1
,ab
(x a)(x b)
28.
1
,a0
x(x a)2
30.
x
,ab
(x a)(x b)
CHAPTER 4 GROUP ACTIVITY Interpolating Polynomials
Given two points in the plane, we can use the point-slope form of the equation of a line to find a polynomial
whose graph passes through these two points. How can we proceed if we are given more than two points? For
example, how can we find the equation of a polynomial P(x) whose graph passes through the points listed in
Table 1 and graphed in Figure 1?
y
TABLE 1
5
x
1
2
3
4
P(x)
1
3
3
1
5
x
5
FIGURE 1
The key to solving this problem is to write the unknown polynomial P(x) in the following special form:
P(x) a0 a1(x 1) a2(x 1)(x 2) a3(x 1)(x 2)(x 3)
(1)
Since the graph of P(x) is to pass through each point in Table 1, we can substitute each value of x in (1) to
determine the coefficients a0, a1, a2, and a3. First we evaluate (1) at x 1 to determine a0:
1 P(1)
a0
All other terms in (1) are 0 when x 1.
Using this value for a0 in (1) and evaluating at x 2, we have
3 P(2) 1 a1(1)
All other terms are 0.
2 a1
Continuing in this manner, we have
3 P(3) 1 2(2) a2(2)(1)
8 2a2
4 a2
346
4 Polynomial and Rational Functions
01 P(4) 1 2(3) 4(3)(2) a3(3)(2)(1)
18 6a3
03 a3
We have now evaluated all the coefficients in (1) and can write
P(x) 1 2(x 1) 4(x 1)(x 2) 3(x 1)(x 2)(x 3)
(2)
If we expand the products in (2) and collect like terms, we can express P(x) in the more conventional form
(verify this)
P(x) 3x3 22x2 47x 27
(A) To check these calculations, evaluate P(x) at x 1, 2, 3, and 4 and compare the results with Table 1. Then
add the graph of P(x) to Figure 1.
(B) Write a verbal description of the special form of P(x) in (1).
In general, given a set of n 1 points:
x
x0
x1
xn
y
y0
y1
yn
the interpolating polynomial for these points is the polynomial P(x) of degree less than or equal to n that satisfies P(xk) yk for k 0, 1, . . . , n. The general form of the interpolating polynomial is
P(x) a0 a1(x x0) a2(x x0)(x x1) an(x x0)(x x1) (x xn1)
(C) Summarize the procedure for using the points in the table to find the coefficients in the general form.
(D) Give an example to show that the interpolating polynomial can have degree strictly less than n.
(E) Could there be two different polynomials of degree less than or equal to n whose graph passes through the
given n 1 points? Justify your answer.
(F) Find the interpolating polynomial for each of Tables 2 and 3. Check your answers by evaluating the polynomial, and illustrate by graphing the points in the table and the polynomial on the same axes.
TABLE 2
TABLE 3
x
1
0
1
2
x
2
1
0
1
2
y
5
3
3
11
y
3
0
5
0
3
A surprisingly short program on a graphing utility can be used to calculate the coefficients in the general
form of an interpolating polynomial. Table 4 shows such a program for a Texas Instruments graphing calculator and the output generated when we use the program to find the coefficients of the interpolating polynomial for Table 1.
Chapter 4 Review
347
TABLE 4 Interpolating Polynomial Coefficients on a Graphing Utility
Program INTERP*
Output
L2→L3
dimL L3→M
For(I,2,M,1)
For(J,M,I,-1)
(L3(J)-L3(J-1))/(L1(J)-L1(J-I+1))→L3(J)
End
End
Disp L3
(G) If you have a TI-85 or TI-86 graphing calculator, enter INTERP in your calculator exactly as shown in
Table 4. To use the program, enter the x values in L1 and the corresponding y values in L2 (see the output in Table 4) and then execute the program. If you have some other graphing utility that can store and
execute programs, consult your manual and modify the statements in INTERP so that the program works
on your graphing utility. Use INTERP to check your answers to part F.
*Available for download at www.mhhe.com/barnett.
Chapter 4 Review
In this chapter, unless indicated otherwise, the coefficients of
the nth-degree polynomial function P(x) anxn an1xn1
a1x a0 are complex numbers and the domain is the
set of complex numbers. The number r is said to be a zero of the
function P, or a zero of the polynomial P(x), or a solution or
root of the equation P(x) 0, if P(r) 0. If the coefficients of
P(x) are real numbers, then the x intercepts of the graph of
y P(x) are real zeros of P and P(x) and real solutions or roots
for the equation P(x) 0.
4-1
POLYNOMIAL FUNCTIONS AND GRAPHS
Synthetic division is an efficient method for dividing polynomials by linear terms of the form x r that is well-suited to calculator use.
Let P(x) be a polynomial of degree greater than 0 and
let r be a real number. Then we have the following important
theorems:
depending on n and the sign of an. A turning point on a continuous graph is a point that separates an increasing portion from a
decreasing portion. Important graph properties are:
1. P is continuous for all real numbers.
2. The graph of P is a smooth curve.
3. The graph of P has at most n x intercepts.
4. P has at most n 1 turning points.
4-2
FINDING RATIONAL ZEROS OF POLYNOMIALS
If P(x) is a polynomial of degree n 0, then we have the following important theorems:
Factor Theorem. The number r is a zero of P(x) if and only if
(x r) is a factor of P(x).
Fundamental Theorem of Algebra. P(x) has at least one zero.
Division Algorithm. P(x) (x r)Q(x) R, where x r is
the divisor; Q(x), a unique polynomial of degree 1 less than
P(x), is the quotient; and R, a unique real number, is the
remainder.
n Zeros Theorem. P(x) can be expressed as a product of n linear factors and has n zeros, not necessarily distinct.
Remainder Theorem. P(r) R.
If P(x) is represented as the product of linear factors and
x r occurs m times, then r is called a zero of multiplicity m.
The left and right behavior of an nth-degree polynomial
P(x) with real coefficients is determined by its highest degree or
leading term. As x → , an xn and P(x) both approach ,
Imaginary Zeros Theorem. If P(x) has real coefficients, then
imaginary zeros of P(x), if they exist, must occur in conjugate pairs.