Copyright © Cengage Learning. All rights reserved. Equations and Inequalities 2.
Copyright © Cengage Learning. All rights reserved. Systems of Equations and Inequalities.
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Transcript of Copyright © Cengage Learning. All rights reserved. Systems of Equations and Inequalities.
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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
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
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
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
We now consider the case in which the denominator has irreducible quadratic factors, some of which are repeated.