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B. Rational Expressions and Partial Fractions

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



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



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B. Rational Expressions and Partial Fractions

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