B. Rational Expressions and Partial Fractions
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3
5
5
3
ϭ
ϩ 2
ϩ
xϪ1
xϪ1
1x Ϫ 12 1x Ϫ 12
x Ϫ 2x ϩ 1
31x Ϫ 12
5
ϭ
ϩ
1x Ϫ 12 1x Ϫ 12
1x Ϫ 12 1x Ϫ 12
13x Ϫ 32 ϩ 5
ϭ
1x Ϫ 121x Ϫ 12
ϭ
3x ϩ 2
1x Ϫ 12 2
683
factor denominators
common denominator
combine numerators
result
Note that while the new denominator is the repeated factor 1x Ϫ 12 2, both 1x Ϫ 12
and 1x Ϫ 12 2 were denominators in the original sum. Assuming we didn’t know the
original sum, reversing the process would require us to begin with the template
3x ϩ 2
A
B
ϭ
ϩ
2
x
Ϫ
1
1x Ϫ 12
1x Ϫ 12 2
and solve for the constants A and B. As with observation 1, we know the numerator of
the first term must be constant. While the second term would still be a proper fraction
if the numerator were linear (degree 1), the denominator is a repeated linear factor and
using a single constant in the numerator of all such fractions will ensure we obtain
unique values for A and B. In the end, for any repeated linear factor 1ax ϩ b2 n in the
A1
A2
AnϪ1
ϩ
ϩpϩ
ϩ
original denominator, terms of the form
2
ax ϩ b
1ax ϩ b2
1ax ϩ b2 nϪ1
An
must appear in the decomposition template, although some of these numer1ax ϩ b2 n
ators may turn out to be zero.
EXAMPLE 3
ᮣ
Writing the Decomposition Template for Unique and Repeated Linear Factors
Write the decomposition template for
xϪ8
xϩ1
a. 2
b. 2
2x ϩ 5x ϩ 3
x Ϫ 6x ϩ 9
Solution
ᮣ
xϪ8
. With two distinct linear
12x ϩ 32 1x ϩ 12
factors in the denominator, the decomposition template is
a. Factoring the denominator gives
A
xϪ8
B
ϭ
ϩ
2x ϩ 3
xϩ1
12x ϩ 321x ϩ 12
decomposition template
xϩ1
, and the denominator is a repeated linear
1x Ϫ 32 2
factor. Using our previous observations the template would be
b. After factoring we have
xϩ1
A
B
ϭ
ϩ
2
xϪ3
1x Ϫ 32
1x Ϫ 32 2
decomposition template
Now try Exercises 23 through 28
ᮣ
When both distinct and repeated linear factors are present in the denominator, the
decomposition template maintains the elements illustrated in both observations 1 and 2.
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EXAMPLE 4
ᮣ
Writing the Decomposition Template for Unique and Repeated Linear Factors
Write the decomposition template for
Solution
ᮣ
Factoring the denominator gives
x2 Ϫ 4x Ϫ 15
.
x3 Ϫ 2x2 ϩ x
x2 Ϫ 4x Ϫ 15
x2 Ϫ 4x Ϫ 15
or
after factoring
x1x2 Ϫ 2x ϩ 12
x1x Ϫ 12 2
completely. With a distinct linear factor of x, and the repeated linear factor 1x Ϫ 12 2,
the decomposition template becomes
A
C
B
x2 Ϫ 4x Ϫ 15
ϭ ϩ
ϩ
2
x
x
Ϫ
1
x1x Ϫ 12
1x Ϫ 12 2
decomposition template
Now try Exercises 29 and 30
ᮣ
To continue our observations,
4
2x ϩ 3
3. Consider the sum ϩ 2
, noting the denominator of the first term is linear,
x
x ϩ1
while the denominator of the second is an irreducible quadratic.
41x2 ϩ 12
12x ϩ 32x
4
2x ϩ 3
ϭ
ϩ 2
ϩ 2
2
x
x ϩ1
x1x ϩ 12
1x ϩ 12x
ϭ
ϭ
14x2 ϩ 42 ϩ 12x2 ϩ 3x2
x1x2 ϩ 12
6x2 ϩ 3x ϩ 4
x1x2 ϩ 12
find common denominator
combine numerators
result
Here, reversing the process would require us to begin with the template
A
6x2 ϩ 3x ϩ 4
Bx ϩ C
ϭ ϩ 2
,
2
x
x1x ϩ 12
x ϩ1
allowing that the numerator of the second term might be linear since the denominator
is quadratic but not due to a repeated linear factor.
1
xϪ2
ϩ 2
4. Finally, consider the sum 2
, where the denominator of the first
x ϩ3
1x ϩ 32 2
term is an irreducible quadratic, with the second being the same factor with multiplicity two.
11x2 ϩ 32
xϪ2
xϪ2
1
ϩ
ϩ 2
ϭ
2
2
2
2
2
x ϩ3
1x ϩ 32
1x ϩ 32 1x ϩ 32
1x ϩ 321x2 ϩ 32
1x2 ϩ 32 ϩ 1x Ϫ 22
ϭ
1x2 ϩ 321x2 ϩ 32
ϭ
WORTHY OF NOTE
Note that the second term in the
decomposition template would still
be a proper fraction if the
numerator were quadratic or cubic,
but since the denominator is a
repeated quadratic factor, using
only a linear form ensures we
obtain unique values for all
coefficients.
x2 ϩ x ϩ 1
1x2 ϩ 32 2
common denominator
combine numerators
result after simplifying
Reversing the process would require us to begin with the template
Cx ϩ D
Ax ϩ B
x2 ϩ x ϩ 1
ϩ 2
ϭ 2
2
2
1x ϩ 32
x ϩ3
1x ϩ 32 2
allowing that the numerator of either term might be nonconstant for the reasons in observation 3. Similar to our reasoning in observation 2, all powers of a repeated quadratic factor must be present in the template.
When both distinct and repeated factors are present in the denominator, the decomposition template maintains the essential elements determined by observations 1
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685
through 4. Using these observations, we can formulate a general approach to the decomposition template.
Decomposition Template for Rational Expressions
For the rational expression
1.
2.
3.
4.
EXAMPLE 5
ᮣ
P1x2
in lowest terms . . .
Q1x2
Factor Q completely into linear factors and irreducible quadratic factors.
For the linear factors, each distinct linear factor and each power of a repeated
linear factor must appear in the decomposition template with a constant
numerator.
For the irreducible quadratic factors, each distinct quadratic factor and each
power of a repeated quadratic factor must appear in the decomposition
template with a linear numerator.
If the degree of P is greater than or equal to the degree of Q, find the quotient
and remainder using polynomial division. Only the remainder portion need
be decomposed into partial fractions.
Writing the Decomposition Template for Linear and Quadratic Factors
Write the decomposition template for
x2 ϩ 10x ϩ 1
x2
a.
b.
1x ϩ 121x2 ϩ 3x ϩ 12
1x2 ϩ 22 3
Solution
ᮣ
a. One factor of the denominator is a distinct linear factor, and the other is an
irreducible quadratic. The decomposition template is
A
x2 ϩ 10x ϩ 1
Bx ϩ C
ϭ
ϩ 2
2
x
ϩ
1
1x ϩ 121x ϩ 3x ϩ 12
x ϩ 3x ϩ 1
decomposition template
b. The denominator consists of a repeated, irreducible quadratic factor. Using our
previous observations the template would be
Cx ϩ D
Ax ϩ B
Ex ϩ F
x2
ϩ 2
ϭ 2
ϩ 2
2
3
2
1x ϩ 22
x ϩ2
1x ϩ 22
1x ϩ 22 3
decomposition template
Now try Exercises 31 and 32
ᮣ
Once the template is obtained, we multiply both sides of the equation by the factored
form of the original denominator and simplify. The resulting equation is an identity—a
true statement for all real numbers x, and in many cases the constants A, B, C, and so on
can be identified using a choice of convenient values for x, as in Example 6.
EXAMPLE 6
ᮣ
Decomposing a Rational Expression with Linear Factors
Decompose the expression
Solution
ᮣ
4x ϩ 11
into partial fractions.
x ϩ 7x ϩ 10
2
4x ϩ 11
, with two distinct linear factors in
1x ϩ 52 1x ϩ 22
the denominator. The required template is
Factoring the denominator gives
A
B
4x ϩ 11
ϭ
ϩ
xϩ5
xϩ2
1x ϩ 52 1x ϩ 22
decomposition template
Multiplying both sides by 1x ϩ 521x ϩ 22 clears all denominators and yields
4x ϩ 11 ϭ A1x ϩ 22 ϩ B1x ϩ 52
clear denominators
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Since the equation must be true for all x, using x ϭ Ϫ5 will conveniently
eliminate the term with B, and enable us to solve for A directly:
41Ϫ52 ϩ 11 ϭ A1Ϫ5 ϩ 22 ϩ B1Ϫ5 ϩ 52
Ϫ20 ϩ 11 ϭ Ϫ3A ϩ B102
Ϫ9 ϭ Ϫ3A
3ϭA
substitute ؊5 for x
simplify
term with B is eliminated
solve for A
To find B, we repeat this procedure, using an x-value that conveniently
eliminates the term with A, namely, x ϭ Ϫ2.
4x ϩ 11 ϭ A1x ϩ 22 ϩ B1x ϩ 52
41Ϫ22 ϩ 11 ϭ A1Ϫ2 ϩ 22 ϩ B1Ϫ2 ϩ 52
Ϫ8 ϩ 11 ϭ A102 ϩ 3B
3 ϭ 3B
1ϭB
original equation
substitute ؊2 for x
simplify
term with A is eliminated
solve for B
With A ϭ 3 and B ϭ 1, the complete decomposition is
3
1
4x ϩ 11
ϭ
ϩ
xϩ5
xϩ2
1x ϩ 52 1x ϩ 22
which can be checked by adding the rational expressions on the right.
Now try Exercises 33 through 38
EXAMPLE 7
ᮣ
Decomposing a Rational Expression with Repeated Linear Factors
Decompose the expression
Solution
ᮣ
ᮣ
9
into partial fractions.
1x ϩ 521x2 ϩ 7x ϩ 102
9
9
ϭ
1x ϩ 52 1x ϩ 221x ϩ 52
1x ϩ 221x ϩ 52 2
(one distinct linear factor, one repeated linear factor). The decomposition template
A
9
B
C
ϭ
ϩ
ϩ
is
. Multiplying both sides by
2
xϩ2
xϩ5
1x ϩ 22 1x ϩ 52
1x ϩ 52 2
Factoring the denominator gives
1x ϩ 221x ϩ 52 2 clears all denominators and yields
9 ϭ A1x ϩ 52 2 ϩ B1x ϩ 22 1x ϩ 52 ϩ C1x ϩ 22.
Using x ϭ Ϫ5 will eliminate the terms with A and B, giving
9 ϭ A1Ϫ5 ϩ 52 2 ϩ B1Ϫ5 ϩ 22 1Ϫ5 ϩ 52 ϩ C1Ϫ5 ϩ 22
9 ϭ A102 ϩ B1Ϫ32102 Ϫ 3C
9 ϭ Ϫ3C
Ϫ3 ϭ C
substitute ؊5 for x
simplify
terms with A and B are eliminated
solve for C
Using x ϭ Ϫ2 will eliminate the terms with B and C, and we have
9 ϭ A1x ϩ 52 2 ϩ B1x ϩ 22 1x ϩ 52 ϩ C1x ϩ 22
9 ϭ A1Ϫ2 ϩ 52 2 ϩ B1Ϫ2 ϩ 22 1Ϫ2 ϩ 52 ϩ C1Ϫ2 ϩ 22
9 ϭ A132 2 ϩ B102 132 ϩ C102
9 ϭ 9A
1ϭA
original equation
substitute ؊2 for x
simplify
terms with B and C are eliminated
solve for A
To find B, we substitute A ϭ 1 and C ϭ Ϫ3 into the original equation, with
any value of x that does not eliminate B. For efficiency, we’ll often use x ϭ 0 or
x ϭ 1 for this purpose (if possible).
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9 ϭ A1x ϩ 52 2 ϩ B1x ϩ 22 1x ϩ 52 ϩ C1x ϩ 22
9 ϭ 110 ϩ 52 2 ϩ B10 ϩ 2210 ϩ 52 Ϫ 310 ϩ 22
9 ϭ 25 ϩ 10B Ϫ 6
Ϫ1 ϭ B
687
original equation
substitute 1 for A, ؊3 for C, 0 for x
simplify
solve for B
With A ϭ 1, B ϭ Ϫ1, and C ϭ Ϫ3 the complete decomposition is
Ϫ1
Ϫ3
9
1
ϩ
ϩ
ϭ
2
xϩ2
xϩ5
1x ϩ 221x ϩ 52
1x ϩ 52 2
1
1
3
ϭ
Ϫ
Ϫ
xϩ2
xϩ5
1x ϩ 52 2
Now try Exercises 39 and 40
ᮣ
As an alternative to using convenient values, a system of equations can be set up
by multiplying out the right-hand side (after clearing fractions) and equating coefficients of the terms with like degrees.
EXAMPLE 8
ᮣ
Decomposing a Rational Expression with Linear and Quadratic Factors
Decompose the given expression into partial fractions:
Solution
ᮣ
3x2 Ϫ x Ϫ 11
.
x3 Ϫ 3x2 ϩ 4x Ϫ 12
A careful inspection indicates the denominator will factor by grouping, giving
x3 Ϫ 3x2 ϩ 4x Ϫ 12 ϭ x2 1x Ϫ 32 ϩ 41x Ϫ 32 ϭ 1x Ϫ 32 1x2 ϩ 42 . With one linear
factor and one irreducible quadratic factor, the required template is
A
Bx ϩ C
3x2 Ϫ x Ϫ 11
ϭ
ϩ 2
2
xϪ3
1x Ϫ 321x ϩ 42
x ϩ4
3x2 Ϫ x Ϫ 11 ϭ A1x2 ϩ 42 ϩ 1Bx ϩ C21x Ϫ 32
ϭ Ax2 ϩ 4A ϩ Bx2 Ϫ 3Bx ϩ Cx Ϫ 3C
ϭ 1A ϩ B2x2 ϩ 1C Ϫ 3B2x ϩ 4A Ϫ 3C
decomposition template
multiply by 1x Ϫ 321x 2 ϩ 42
(clear denominators)
distribute/F-O-I-L
group and factor
For the left side to equal the right, we must equate coefficients of terms with like
degree: A ϩ B ϭ 3, C Ϫ 3B ϭ Ϫ1, and 4A Ϫ 3C ϭ Ϫ11. This gives the 3 ϫ 3
1
1
0
3
system £ 0 Ϫ3
1
Ϫ1 § in matrix form (verify this). Using the matrix of
4
0
Ϫ3 Ϫ11
coefficients we find that D ϭ 13 (see figure), and we complete the solution using
Cramer’s rule.
3
DA ϭ † Ϫ1
Ϫ11
1
0
1
Ϫ3 1 † ϭ 13 DB ϭ † 0
0 Ϫ3
4
3
0
1
Ϫ1
1 † ϭ 26 DC ϭ † 0
Ϫ11 Ϫ3
4
1
3
Ϫ3 Ϫ1 † ϭ 65
0 Ϫ11
26
65
The result is A ϭ 13
13 ϭ 1, B ϭ 13 ϭ 2, and C ϭ 13 ϭ 5, giving the decomposition
1
2x ϩ 5
3x2 Ϫ x Ϫ 11
ϭ
ϩ 2
.
2
xϪ3
1x Ϫ 321x ϩ 42
x ϩ4
Now try Exercises 41 through 46
ᮣ
In some cases, the “convenient values” method cannot be applied and a system of
equations is our only option. Also, if the decomposition template produces a large or
cumbersome system, a graphing calculator can assist the solution process using a
matrix equation or the “rref(” feature. See Exercises 47 and 48.
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As a final reminder, if the degree of the numerator is greater than the degree of the
denominator, divide using long division and apply the preceding methods to the remainder
2x Ϫ 7
3x3 ϩ 6x2 ϩ 5x Ϫ 7
ϭ 3x ϩ
, and
polynomial. For instance, you can check that
2
x ϩ 2x ϩ 1
1x ϩ 12 2
9
2
Ϫ
decomposing the remainder polynomial gives a final result of 3x ϩ
.
xϩ1
1x ϩ 12 2
B. You’ve just seen how
we can decompose a rational
expression into partial
fractions
C. Determinants, Geometry, and the Coordinate Plane
As mentioned in the introduction, the use of determinants extends far beyond solving
systems of equations. Here, we’ll demonstrate how determinants can be used to find
the area of a triangle whose vertices are given as three points in the coordinate plane.
The Area of a Triangle in the xy-Plane
Given a triangle with vertices at (x1, y1), (x2, y2), and (x3, y3),
Area ϭ `
EXAMPLE 9
ᮣ
det1T2
2
x1
` where T ϭ £ x2
x3
y1
y2
y3
1
1§
1
Finding the Area of a Triangle Using Determinants
Find the area of a triangle with vertices at (3, 1), (Ϫ2, 3), and (1, 7) (see Figure 7.25).
Solution
ᮣ
Figure 7.25
y
7
2
(1, 7)
6
5
4
(Ϫ2, 3)
3
2
(3, 1)
1
Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1
Ϫ1
1
2
3
Begin by forming matrix T and computing det(T) (see Figure 7.26):
Figure 7.26
3 1 1
x1 y1 1
T ϭ [A]
det1T2 ϭ † x y 1 † ϭ † Ϫ2 3 1 †
4
5
x
Ϫ2
Ϫ3
2
x3 y3 1
1 7 1
ϭ 313 Ϫ 72Ϫ11Ϫ2 Ϫ 12 ϩ 11Ϫ14 Ϫ 32
ϭ Ϫ12 ϩ 3 ϩ 1Ϫ172 ϭ Ϫ26
det1T 2
Ϫ26
` ϭ `
`
Compute the area: A ϭ `
2
2
ϭ 13
The area of this triangle is 13 units2.
Now try Exercises 51 through 56
ᮣ
As an extension of this formula, what if the three points were collinear? After a
moment, it may occur to you that the formula would give an area of 0 units2, since no
triangle could be formed. This gives rise to a test for collinear points.
Test for Collinear Points
Three points (x1, y1), (x2, y2), and (x3, y3) are collinear if
C. You’ve just seen how we
can use determinants in
applications involving geometry
in the coordinate plane
x1
det1A2 ϭ † x2
x3
y1
y2
y3
1
1 † ϭ 0.
1
See Exercises 57 through 62. There are a variety of additional applications in the
Exercise Set. See Exercises 63 through 68.
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7.4 EXERCISES
ᮣ
CONCEPTS AND VOCABULARY
Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.
1. The determinant `
as:
2.
a11
a21
a12
` is evaluated
a22
rule uses a ratio of determinants to
solve for the unknowns in a system.
x1
y1
1
x3
y3
1
collinear if 0T 0 ϭ † x2 y2 1 † has a value of
.
3. Given the matrix of coefficients D, the matrix Dx is
formed by replacing the coefficients of x with the
terms.
ᮣ
4. The three points (x1, y1), (x2, y2), and (x3, y3) are
5. Discuss/Explain the process of writing
.
8x Ϫ 3
as a
x2 Ϫ x
sum of partial fractions.
6. Discuss/Explain why Cramer’s rule cannot be
applied if D ϭ 0. Use an example to illustrate.
DEVELOPING YOUR SKILLS
Write the determinants D, Dx, and Dy for the systems
given, but do not solve.
7. e
2x ϩ 5y ϭ 7
Ϫ3x ϩ 4y ϭ 1
8. e
Ϫx ϩ 5y ϭ 12
3x Ϫ 2y ϭ Ϫ8
Solve each system of equations using Cramer’s rule, if
possible. Do not use a calculator.
9. e
4x ϩ y ϭ Ϫ11
3x Ϫ 5y ϭ Ϫ60
10. e
x ϭ Ϫ2y Ϫ 11
y ϭ 2x Ϫ 13
y
x
ϩ ϭ1
8
4
11. μ
y
x
ϭ ϩ6
5
2
3
7
2
xϪ yϭ
3
8
5
12. μ
5
3
11
xϩ yϭ
6
4
10
13. e
0.6x Ϫ 0.3y ϭ 8
0.8x Ϫ 0.4y ϭ Ϫ3
14. e
Ϫ2.5x ϩ 6y ϭ Ϫ1.5
0.5x Ϫ 1.2y ϭ 3.6
The two systems given in Exercises 15 and 16 are identical except for the third equation. For the first system given,
(a) write the determinants D, Dx, Dy, and Dz then (b) determine if a solution using Cramer’s rule is possible by
computing ͦ Dͦ without the use of a calculator (do not solve the system). Then (c) compute ͦDͦ for the second system
and try to determine how the equations in the second system are related.
4x Ϫ y ϩ 2z ϭ Ϫ5
4x Ϫ y ϩ 2z ϭ Ϫ5
15. • Ϫ3x ϩ 2y Ϫ z ϭ 8 , • Ϫ3x ϩ 2y Ϫ z ϭ 8
x Ϫ 5y ϩ 3z ϭ Ϫ3
xϩ yϩ zϭ3
2x ϩ
3z ϭ Ϫ2
2x ϩ
3z ϭ Ϫ2
16. • Ϫx ϩ 5y ϩ z ϭ 12 , • Ϫx ϩ 5y ϩ z ϭ 12
3x Ϫ 2y ϩ z ϭ Ϫ8
x ϩ 5y ϩ 4z ϭ 10
Use Cramer’s rule to solve each system of equations. Verify computations using a graphing calculator.
x ϩ 2y ϩ 5z ϭ 10
17. • 3x ϩ 4y Ϫ z ϭ 10
x Ϫ y Ϫ z ϭ Ϫ2
x ϩ 3y ϩ 5z ϭ 6
18. • 2x Ϫ 4y ϩ 6z ϭ 14
9x Ϫ 6y ϩ 3z ϭ 3
y ϩ 2z ϭ 1
19. • 4x Ϫ 5y ϩ 8z ϭ Ϫ8
8x Ϫ 9z ϭ 9
x ϩ 2y ϩ 5z ϭ 10
20. • 3x Ϫ z ϭ 8
Ϫy Ϫ z ϭ Ϫ3
w ϩ 2x Ϫ 3y ϭ Ϫ8
x Ϫ 3y ϩ 5z ϭ Ϫ22
21. μ
4w Ϫ 5x ϭ 5
Ϫy ϩ 3z ϭ Ϫ11
w Ϫ 2x ϩ 3y Ϫ z ϭ 11
3w Ϫ 2y ϩ 6z ϭ Ϫ13
22. μ
2x ϩ 4y Ϫ 5z ϭ 16
3x Ϫ 4z ϭ 5