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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
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)
= (A + B + C)x2 + (3A + B)x + (2A – 2B – C)
If two polynomials are equal, then their coefficients are
equal. Thus since 5x + 7 has no x2-term, we have
A + B + C = 0.
cont’d
Expand
Combine
like terms
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Example 1 – Solution
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
cont’d
Equation 1: Coefficients of x2
Equation 2: Coefficients of x
Equation 3: Constant coefficients
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Example 1 – Solution
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
cont’d
Equation 2 + (–3) Equation 1
Equation 3 + (–2) Equation 1
Equation 3 + (–2) Equation 2
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Example 1 – Solution
From the third equation we get C = –1. Back-substituting,
we find that B = –1 and A = 2.
So the partial fraction decomposition is
cont’d
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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
= A(x3 – 3x2 + 3x – 1) + B(x3 – 2x2 + x) + C(x2 – x) + Dx
= (A + B)x3 + (–3A – 2B + C)x2 + (3A + B – C + D)x – A
Equating coefficients, we get the following equations.
A + B = 0
–3A – 2B + C = 5
3A + B – C + D = 7
–A = –1
cont’d
Expand
Combine like terms
Coefficients of x3
Coefficients of x2
Coefficients of x
Constant coefficients
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Example 2 – Solution
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
cont’d
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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
Equating coefficients gives us the equations
A + B = 2
C = –1
4A = 4
so A = 1, B = 1, and C = –1. The required partial fraction
decomposition is
cont’d
Coefficients of x2
Coefficients of x
Constant coefficients
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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:
