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)
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,
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, I1 and I2, with I1 going through one resistor and I2 going through the other, so that I=I1+I2. Suppose we want to find the total resistance R in the circuit due to the two resistors R1 and R2 connected in parallel. Ohm’s law can be used to show that in the parallel circuit in Figure 5.9, 1R=1R1+1R2. This result is called the parallel property of resistors. In Example 8, we examine the parallel property of resistors by using partial fractions.
1 Become familiar with partial-fraction decomposition.
Recall that to add two rational expressions such as 2x+3 and 3x−1 required you to find the least common denominator, rewrite expressions with the same denominator, and then add the corresponding numerators. This procedure gives
In some applications of algebra and more advanced mathematics, you need a reverse procedure for splitting a fraction such as 5x+7(x−1)(x+3) into the simpler fractions to obtain:
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 P(x)Q(x) is called improper if the degree P(x)≥degree Q(x) and is called proper if the degree P(x)<degree Q(x). Examples of improper rational expressions are
Using long division, we can express an improper rational fraction P(x)Q(x) as the sum of a polynomial and a proper rational fraction:
We therefore restrict our discussion of the decomposition of 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 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).
2 Decompose P(x)Q(x) when Q(x) has only distinct linear factors.
We can find the constants A1, A2, …, An by using the following procedure. Note that if the number of constants is small, we use the letters A, B, C, …, instead of A1, A2, A3, ….
Find the partial-fraction decomposition of 2x−7(x+1)(x−2).
Find the partial-fraction decomposition of the expression
First, factor the denominator.
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
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 10x−4=0x2+10x−4, we have
Equating corresponding coefficients leads to the system of equations.
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
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:
Substitute x=2 in equation (1) to cause the terms containing A and C to be 0.
Now substitute x=−2 in equation (1) to get
Finally, substitute x=0 in equation (1) to get
Because we found the same values for A, B, and C, the partial-fraction decomposition will also be the same.
Find the partial-fraction decomposition of
3 Decompose P(x)Q(x) when Q(x) has repeated linear factors.
Find the partial-fraction decomposition of x+4(x+3)(x−1)2.
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:
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:
Substitute x=1 in equation (1) to obtain:
This simplifies to 5=4C, so C=54.
To find A, we substitute x=−3 in equation (1) to get
This simplifies to 1=16A, so A=116.
To find B, we replace x with any convenient number, say, 0 in equation (1). We have
Now replace A with 116 and C with 54 in equation (2) to get:
We have A=116,B=−116, and C=54.
Step 5 Substitute A=116, B=−116, and C=54 in the decomposition in Step 1 to get
or
Find the partial-fraction decomposition of
Find the partial-fraction decomposition of 2x2−7x+9(x−1)3.
Step 1 The only linear factor of the denominator, x−1, is repeated 3 times. The partial-fraction decomposition has the form:
Step 2 Multiply both sides of equation (1) by (x−1)3 and simplify to obtain the identity:
Step 3 Substitute x=1 in equation (2) to get C=4.
So, equation (2) becomes:
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.
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
Find the partial-fraction decomposition of 2x+1(x+3)2.
4 Decompose P(x)Q(x) when Q(x) has distinct irreducible quadratic factors.
Find the partial-fraction decomposition of 3x2−8x+1(x−4)(x2+1).
Step 1 The factor x−4 is linear and x2+1 is irreducible; so the partial-fraction decomposition has the form:
Step 2 Multiply both sides of the decomposition in Step 1 by the original denominator, (x−4)(x2+1), and simplify to get
Substitute x=4 to find 17=17A, or A=1.
Step 3 Collect like terms and write both sides of the equation in Step 2 in descending powers of x.
Equate corresponding coefficients to get
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 1−4C=1, or C=0.
Step 5 Substitute A=1, B=2, and C=0 into the decomposition in Step 1 to get:
Find the partial-fraction decomposition of
5 Decompose P(x)Q(x) when Q(x) has repeated irreducible quadratic factors.
Find the partial-fraction decomposition of 2x4−x3+13x2−2x+13(x−1)(x2+4)2.
Step 1 The denominator has a nonrepeating linear factor (x−1) and an irreducible quadratic factor (x2+4) that appears twice. The decomposition therefore has the form:
Step 2 Multiply both sides of the decomposition in Step 1 by the original denominator (x−1)(x2+4)2, use the distributive property, and eliminate common factors:
Substitute x=1 and simplify to obtain 25=25A, 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
Equate corresponding coefficients to get the system:
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.
Find the partial-fraction decomposition of
Express the sum
as a fraction of whole numbers in lowest terms.
Each term in the sum is of the form: 1k(k+1). From the partial-fraction decomposition, we have:
Therefore, by using this decomposition for each term, we get:
After removing parentheses, we see that most terms cancel in pairs (such as −13 and 13), leaving only the first and the last term. The original sum is therefore referred to as a telescoping sum that “collapses” to
Express the sum 14⋅5+15⋅6+16⋅7+⋯+13111⋅3112 as a fraction of whole numbers in lowest terms.
The total resistance R due to two resistors connected in parallel is given by
Use partial-fraction decomposition to write 1R in terms of partial fractions. What does each term represent?
From R=x(x+5)3x+10, we get the following:
Substitute x=0 in the last equation to obtain 5A=10, or A=2. Now substitute x=−5 in the same equation to get −5=−5B, or B=1. Thus, we have
The last equation says that if two resistors with resistances R1=x2 and R2=x+5 are connected in parallel, they will produce a total resistance R, where R=x(x+5)3x+10.
Repeat Example 8, assuming that R=(x+1)(x+2)4x+7.
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)=x3−9x 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 P(x)Q(x).
True or False. The quadratic expression x2+2x+1 is irreducible.
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.
1(x−1)(x+2)
x(x+1)(x−3)
1x2+7x+6
3x2−6x+8
2x3−x2
x−1(x+2)2(x−3)
x2−3x+3(x+1)(x2−x+1)
2x+3(x−1)2(x2+x+1)
3x−4(x2+1)2
x−2(2x+3)2(x2+3)2
In Exercises 19–66, find the partial-fraction decomposition of each rational expression.
2x+1(x+1)(x+2)
7(x−2)(x+5)
1x2+4x+3
xx2+5x+6
2x2+2x
x+9x2−9
x(x+1)(x+2)(x+3)
8x−14(x−1)(x+2)(x−3)
x2(x−1)(x2+5x+4)
2x+14(x+1)(x2−6x+5)
3x+1(x−5)(x+3)2
1−2x(x+3)(x−4)2
8x+1(x+2)2(x−3)
3x+2x2(x+1)
x−1(x+1)2
x(x−2)2
x−1(2x−3)2
6x+1(3x+1)2
2x2+x(x+1)3
x2+2(x−1)3
2x+3(x+2)3
x2−8x+18(x−5)3
−x2+3x+1x3+2x2+x
5x2−8x+2x3−2x2+x
3x+1(x+1)(x2−1)
6x−4(x−2)(x2−4)
1x2(x+1)2
2(x−1)2(x+3)2
1(x2−1)2
3(x2−4)2
x2−5(x−2)(x2−2x+3)
x2+2x+1(x+2)(x2+x+2)
x−1(x+1)2(x2+2x+2)
x2−3(x−2)2(x2−x+1)
x2+1(x−2)(x2−3x+2)
x+1x(2x2−5x−3)
6x+74x2+12x+9
2x+39x2+30x+25
x−3x3+x2
x2+1x4−x3
x2+2x+4x3+x2
x2+2x−1(x−1)(x2+1)
xx4−1
3x3−5x2+12x+4x4−16
1x(x2+1)2
x2(x2+2)2
2x2+3x(x2+1)(x2+2)
x2−2x(x2+9)(x2+x+7)
In Exercises 67–70, express each sum as a fraction of whole numbers in lowest terms.
11⋅2+12⋅3+13⋅4+⋯+ 1n(n+1)
21⋅3+22⋅4+23⋅5+⋯+ 2100⋅102
21⋅3+23⋅5+25⋅7+⋯+ 2(2n−1)(2n+1)
21⋅2⋅3+22⋅3⋅4+23⋅4⋅5+⋯+ 2100⋅101⋅102
[Hint: Write2k(k+1)(k+2)=1k−2k+1+1k+2=1k−1k+1−1k+1+1k+2. ]
In Exercises 71–74, the total resistance R in a circuit is given. Use partial-fraction decomposition to write 1R in terms of partial fractions. Interpret each term in the partial fraction.
Circuit resistance. R=(x+1)(x+3)2x+4
Circuit resistance. R=(x+2)(x+4)7x+20
Circuit resistance. R=R1R2R3R1R2+R2R3+R3R1
Circuit resistance. R=x(x+2)(x+4)3x2+12x+8
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.
1x3+1
4x(x2−1)2
2x+3(x2+1)(x+1)
x+1(x2+1)(x−1)2
x(x2+1)2(x−1)
x3(x+1)2(x+2)2
Find the partial-fraction decomposition of
[Hint: x4+4=x4+4+4x2−4x2=(x2+2)2−(2x)2=(x2−2x+2)(x2+2x+2).]
In Exercises 83 and 84, find the partial-fraction decomposition by first using long division.
x2+4x+5x2+3x+2
(x−1)(x+2)(x+3)(x−4)
Show that 1√3+√5+1√5+√7+1√7+√9+⋯+1√2n+1+√2n+3=√2n+3−√32.
Show that 11⋅3⋅5+13⋅5⋅7+15⋅7⋅9+⋯+1(2n−1)(2n+1)(2n+3)=n(n+2)3(2n+1)(2n+3).
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)=x+8x2+x−2=g(x)+h(x), where g(x) and h(x) are the terms in the partial-fraction decomposition of f(x).
Sketch the graphs of f, g, and h.
Compare the graphs of f and g. What do you observe?
Compare the graphs of f and h. What do you observe?
In Exercises 89–92, solve each equation for x.
x2+10x+21=0
x2−2x−24=0
2x2+x−10=0
6x2+x−2=0
In Exercises 93–96, solve each system of equations by the substitution method.
{x+y=33x+2y=7
{3x+y=−34x+3y=1
{x−2y=12x+3y=16
{x−2y=1012x+13y=1
In Exercises 97–100, solve each system of equations by the elimination method.
{2x−y=43x+2y=13
{2x+3y=−13x−2y=5
{2x+5y=13x−2y=−8
{2x−3y=6−3x+2y=1