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Systems of Equations and
Inequalities
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10.7
Partial Fractions
Copyright © Cengage Learning. All rights reserved.
Objectives
► Distinct Linear Factors
► Repeated Linear Factors
► Irreducible Quadratic Factors
► Repeated Irreducible Quadratic Factors
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Partial Fractions
To write a sum or difference of fractional expressions as a
single fraction, we bring them to a common denominator.
For example,
But for some applications of algebra to calculus we must
reverse this process—that is, we must express a fraction
such as 3x/(2x2 – x – 1) as the sum of the simpler fractions
1/(x – 1) and 1/(2x + 1).
These simpler fractions are called partial fractions; we learn
how to find them in this section.
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Partial Fractions
Let r be the rational function
r(x) =
where the degree of P is less than the degree of Q.
By the Linear and Quadratic Factors Theorem, every
polynomial with real coefficients can be factored completely
into linear and irreducible quadratic factors, that is, factors
of the form ax + b and ax2 + bx + c, where a, b, and c are
real numbers.
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Partial Fractions
For instance,
x4 – 1 = (x2 – 1) (x2 + 1) = (x – 1)(x + 1)(x2 + 1)
After we have completely factored the denominator Q of r,
we can express r(x) as a sum of partial fractions of the
form
and
This sum is called the partial fraction decomposition of r.
Let’s examine the details of the four possible cases.
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Distinct Linear Factors
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Distinct Linear Factors
We first consider the case in which the denominator factors
into distinct linear factors.
The constants A1, A2, . . . , An are determined as in the next
example.
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Example 1 – Distinct Linear Factors
Find the partial fraction decomposition of
.
Solution:
The denominator factors as follows.
x3 + 2x2 – x – 2 = x2(x + 2) – (x + 2)
= (x2 – 1) (x + 2)
= (x – 1) (x + 1) (x + 2)
This gives us the partial fraction decomposition
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Example 1 – Solution
cont’d
Multiplying each side by the common denominator,
(x – 1)(x + 1)(x + 2), we get
5x + 7 = A(x + 1)(x + 2) + B(x – 1)(x + 2) + C(x – 1)(x + 1)
= A(x2 + 3x + 2) + B(x2 + x – 2) + C(x2 – 1)
Expand
= (A + B + C)x2 + (3A + B)x + (2A – 2B – C)
Combine
like terms
If two polynomials are equal, then their coefficients are
equal. Thus since 5x + 7 has no x2-term, we have
+ B + C = 0.
A
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Example 1 – Solution
cont’d
Similarly, by comparing the coefficients of x, we see that
3A + B = 5, and by comparing constant terms, we get
2A – 2B – C = 7.
This leads to the following system of linear equations for A,
B, and C.
A+B+C=0
3A + B = 5
2A – 2B – C = 7
Equation 1: Coefficients of x2
Equation 2: Coefficients of x
Equation 3: Constant coefficients
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Example 1 – Solution
cont’d
We use Gaussian elimination to solve this system.
A+ B + C = 0
– 2B – 3C = 5
– 4B – 3C = 7
A + B +C = 0
– 2B – 3C = 5
3C = –3
Equation 2 + (–3)  Equation 1
Equation 3 + (–2)  Equation 1
Equation 3 + (–2)  Equation 2
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Example 1 – Solution
cont’d
From the third equation we get C = –1. Back-substituting,
we find that B = –1 and A = 2.
So the partial fraction decomposition is
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Repeated Linear Factors
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Repeated Linear Factors
We now consider the case in which the denominator
factors into linear factors, some of which are repeated.
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Example 2 – Repeated Linear Factors
Find the partial fraction decomposition of
.
Solution:
Because the factor x – 1 is repeated three times in the
denominator, the partial fraction decomposition has the
form
Multiplying each side by the common denominator,
x(x – 1)3, gives
x2 + 1 = A(x – 1)3 + Bx(x – 1)2 + Cx(x – 1) + Dx
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Example 2 – Solution
cont’d
= A(x3 – 3x2 + 3x – 1) + B(x3 – 2x2 + x) + C(x2 – x) + Dx
Expand
= (A + B)x3 + (–3A – 2B + C)x2 + (3A + B – C + D)x – A
Combine like terms
Equating coefficients, we get the following equations.
A+B=0
–3A – 2B + C = 5
3A + B – C + D = 7
–A = –1
Coefficients of x3
Coefficients of x2
Coefficients of x
Constant coefficients
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Example 2 – Solution
cont’d
If we rearrange these equations by putting the last one in
the first position, we can easily see (using substitution) that
the solution to the system is A = –1, B = 1, C = 0, D = 2, so
the partial fraction decomposition is
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Irreducible Quadratic Factors
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Irreducible Quadratic Factors
We now consider the case in which the denominator has
distinct irreducible quadratic factors.
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Example 3 – Distinct Quadratic Factors
Find the partial fraction decomposition of
.
Solution:
Since x3 + 4x = x(x2 + 4), which can’t be factored further,
we write
Multiplying by x(x2 + 4), we get
2x2 – x + 4 = A(x2 + 4) + (Bx + C)x
= (A + B)x2 + Cx + 4A
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Example 3 – Solution
cont’d
Equating coefficients gives us the equations
A +B=2
C = –1
4A = 4
Coefficients of x2
Coefficients of x
Constant coefficients
so A = 1, B = 1, and C = –1. The required partial fraction
decomposition is
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Repeated Irreducible Quadratic
Factors
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Repeated Irreducible Quadratic Factors
We now consider the case in which the denominator has
irreducible quadratic factors, some of which are repeated.
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Example 4 – Repeated Quadratic Factors
Write the form of the partial fraction decomposition of
Solution:
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