College Algebra

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• College Algebra
• Sixth Edition
• James Stewart ? Lothar Redlin ? Saleem Watson

2

• Systems of Equations and Inequalities

5
3
5.3

• Partial Fractions

4
Introduction

• To write a sum or difference of fractional
  expressions as a single fraction, we bring them
  to a common denominator.

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Introduction

• However, for some applications of algebra to
  calculus, we must reverse this process.
• 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)

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Partial Fractions

• These simpler fractions are called partial
  fractions.
• In this section, we learn how to find them.

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Partial Fractions

• Let r be the rational functionwhere the
  degree of P is less than the degree of Q.

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Partial Fractions

• By the Linear and Quadratic Factors Theorem in
  Section 3.6, 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)

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Partial Fraction Decomposition

• After we have completely factored the
  denominator Q of r, we can express r(x) as a sum
  of partial fractions of the form
• This sum is called the partial fraction
  decomposition of r.

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Partial Fraction Decomposition

• Lets examine the details of the four possible
  cases.

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• Distinct Linear Factors

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Distinct Linear Factors

• The denominator is
• A product of distinct linear factors.

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Distinct Linear Factors

• Suppose that we can factor Q(x) as Q(x)
  (a1x b1 )(a2x b2 ) (anx bn )with no
  factor repeated.
• The partial fraction decomposition of P(x)/Q(x)
  takes the form

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Distinct Linear Factors

• The constants A1, A2, . . . , An are
  determined as in the following example.

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E.g. 1Distinct Linear Factors

• Find the partial fraction decomposition of

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E.g. 1Distinct Linear Factors

• The denominator factors as x3 2x2 x 2
  x2(x 2) (x 2)
  (x2 1)(x 2) (x
  1)(x 1)(x 2)
• This gives the partial fraction decomposition

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E.g. 1Distinct Linear Factors

• Multiplying each side by the common denominator,
  (x 1)(x 1)(x 2), we get5x 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)

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E.g. 1Distinct Linear Factors

• If two polynomials are equal, then their
  coefficients are equal.
• Thus, since 5x 7 has no x2-term, we have A B
  C 0.
• Similarly, by comparing the coefficients of x,
  we see that 3A B 5
• By comparing constant terms, we get 2A
  2B C 7

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E.g. 1Distinct Linear Factors

• This leads to the following system of linear
  equations for A, B, and C.
• We use Gaussian elimination to solve this system.

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E.g. 1Distinct Linear Factors
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E.g. 1Distinct Linear Factors

• 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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Distinct Linear Factors

• The same approach works in the remaining cases.
• We set up the partial fraction decomposition
  with the unknown constants, A, B, C, . . . .
• Then, we multiply each side of the resulting
  equation by the common denominator, simplify the
  right-hand side of the equation, and equate
  coefficients.

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Distinct Linear Factors

• This gives a set of linear equations that will
  always have a unique solution.
• This is provided that the partial fraction
  decomposition has been set up correctly.

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• Repeated Linear Factors

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Repeated Linear Factors

• The denominator is
• A product of linear factors, some of which are
  repeated.

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Repeated Linear Factors

• Suppose the complete factorization of Q(x)
  contains the linear factor ax b repeated k
  timesthat is, (ax b)k is a factor of Q(x).
• Then, corresponding to each such factor, the
  partial fraction decomposition for P(x)/Q(x)
  contains

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E.g. 2Repeated Linear Factors

• Find the partial fraction decomposition of
• The factor x 1 is repeated three times in the
  denominator.

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E.g. 2Repeated Linear Factors

• So, the partial fraction decomposition has the
  form

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E.g. 2Repeated Linear Factors

• We then multiply each side by the common
  denominator x(x 1)3.
• x2 1 A(x 1)3 Bx(x 1)2 Cx(x 1)
  Dx 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

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E.g. 2Repeated Linear Factors

• Equating coefficients, we get
• If we rearrange these 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

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E.g. 2Repeated Linear Factors

• So, the partial fraction decomposition is

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• Irreducible Quadratic Factors

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Irreducible Quadratic Factors

• The denominator has
• Irreducible quadratic factors, none of which is
  repeated.

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Irreducible Quadratic Factors

• Suppose the complete factorization of Q(x)
  contains the quadratic factor ax2 bx c (which
  cant be factored further).
• Then, corresponding to this, the partial fraction
  decomposition of P(x)/Q(x) will have a term of
  the form

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E.g. 3Distinct Quadratic Factors

• Find the partial fraction decomposition of
• Since x3 4x x(x2 4), which cant be
  factored further, we write

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E.g. 3Distinct Quadratic Factors

• 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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E.g. 3Distinct Quadratic Factors

• Equating coefficients gives us
• So, A 1, B 1, and C 1.

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E.g. 3Distinct Quadratic Factors

• The required partial fraction decomposition is

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• Repeated Irreducible Quadratic Factors

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Repeated Irreducible Quadratic Factors

• The denominator has
• A repeated irreducible quadratic factor.

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Repeated Irreducible Quadratic Factors

• Suppose the complete factorization of Q(x)
  contains the factor (ax2 bx c)k, where ax2
  bx c cant be factored further.
• Then, the partial fraction decomposition of
  P(x)/Q(x) will have the terms

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E.g. 4Repeated Quadratic Factors

• Write the form of the partial fraction
  decomposition of

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E.g. 4Repeated Quadratic Factors

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Repeated Quadratic Factors

• To find the values of A, B, C, D, E, F, G, H,
  I, J, and K in Example 4, we would have to solve
  a system of 11 linear equations.
• Although possible, this would certainly involve
  a great deal of work!

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Using Long Division

• The techniques that we have described in this
  section apply only to
• Rational functions P(x)/Q(x) in which the degree
  of P is less than the degree of Q.
• If this isnt the case, we must first use long
  division to divide Q into P.

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E.g. 5Using Long Division to Prepare for Partial
fractions

• Find the partial fraction decomposition of
• The degree of the numerator is larger than the
  degree of the denominator.

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E.g. 5Using Long Division to Prepare for Partial
fractions

• we use long division to obtain
• The remainder term now satisfies the requirement
  that the degree of the numerator is less than
  that of the denominator.

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E.g. 5Using Long Division to Prepare for Partial
fractions

• At this point, we proceed as in Example 1 to
  obtain the decomposition