M1410704Part1

M1410704Part1 - (Section 7.4: Partial Fractions) 7.14...

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(Section 7.4: Partial Fractions) 7.14 SECTION 7.4: PARTIAL FRACTIONS PART A: INTRO A , B , C , etc. represent unknown real constants. Assume that our polynomials have real coefficients. These Examples deal with rational expressions in x , but the methods here extend to rational expressions in y , t , etc. Review how to add and subtract rational expressions in Section A.4: pp.A38-A39 . Review Example 3 x 4 2 x + 1 = 3 x + 1 ( ) Think: "Who's missing?"  2 x 4 ( ) x 4 ( ) x + 1 ( ) = x + 11 x 2 3 x 4 How do we reverse this process? In other words, how do we find that the partial fraction decomposition (PFD) for x + 11 x 2 3 x 4 is 3 x 4 A partial fraction (PF)  2 x + 1 ? The PFD Form that we need depends on the factored form of the denominator . Here, the denominator is x 2 3 x 4 , which factors as x 4 ( ) x + 1 ( ) .
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(Section 7.4: Partial Fractions) 7.15 PART B: THE BIG PICTURE You may want to come back to this Part and Part C after you read the Examples starting on Notes 7.22 . In Section 2.5 , we discussed: “Factoring Over R ” Theorem Let f x ( ) be a nonconstant polynomial in standard form with real coefficients. A complete factorization of f x ( ) over R consists of: 1) Linear factors, 2) Quadratic factors that are R -irreducible (see Note below), or 3) Some product of the above, possibly including repeated factors, and 4) Maybe a nonzero constant factor. Note: A quadratic factor is R -irreducible It has no real zeros (or “roots”), and it cannot be nontrivially factored and broken down further over R (i.e., using only real coefficients). Knowing how to factor such an f x ( ) may pose a problem, however! Finding real zeros of f x ( ) can help you factor f x ( ) ; remember the Factor Theorem from Sections 2.2 and 2.3: Notes 2.19 and 2.33 . Example The hideous polynomial 6 x 14 + 33 x 13 + 45 x 12 + 117 x 11 213 x 10 2076 x 9 3180 x 8 15,024 x 7 11,952 x 6 32,832 x 5 18,240 x 4 19,968 x 3 9216 x 2 can factor over R as follows: 3 x 2 x 3 ( ) 2 x + 1 ( ) x + 4 ( ) 2 x 2 + 1 ( ) x 2 + 4 ( ) 3
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(Section 7.4: Partial Fractions) 7.16 Technical Note: There are other factorizations over R involving manipulations (like “trading”) of constant factors, but we like the fact that the one provided is a factorization over Z (the integers), and we have no factors like 6 x + 3 ( ) for which nontrivial GCFs (greatest common factors) can be pulled out. Let’s categorize factors in this factorization: • 3 is a constant factor. • We will discuss x 2 last. x 3 ( ) is a distinct linear factor. It is distinct (“different”) in the sense that there are no other x 3 ( ) factors, nor are there constant multiples such as 2 x 6 ( ) . 2 x + 1 ( ) is a distinct linear factor. x + 4 ( ) 2 is a [nice] power of a linear factor. Because it can be rewritten as x + 4 ( ) x + 4 ( ) , it is an example of repeated linear factors.
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This note was uploaded on 09/08/2011 for the course MATH 141 taught by Professor Staff during the Fall '11 term at Mesa CC.

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M1410704Part1 - (Section 7.4: Partial Fractions) 7.14...

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