Engng Math 145 Partial Fractions

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11 Terms

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Degree of numerator < denominator

No long division required and partial fractions can be applied immediately.

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Polynomial long division

If degree of numerator greater than degree of denominator, then do long division.

  • Divide dividend by the x term if divisor and write quotient

  • Multiply quotient by the same x term and subtract it from dividend

  • Add dividend of term one lower than the quotient’s current degree

  • Repeat process but on the “difference + dividend term” until degree of the difference term is lower than the degree of the divisor

  • Remainder is remainder divided by the divisor

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Polynomial long division shortcut for equal degrees

Divide leading terms for quotient (y) and then subtract dividend by divisor times quotient (y) for remainder.

eg. (2x2 + 2x + 1) / (x2 + x + 1) → 2x2 / x2 = 2, (2x2 + 2x + 1) - 2(x2 + x + 1) = -1 2 + (-1) / (x2 + x + 1) [answer]

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Distinct linear factors

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Repeated linear factors

A1, A2, …, are unique constants A, B, C, D, …

<p>A<sub>1</sub>, A<sub>2</sub>, …, are unique constants A, B, C, D, …</p>
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Distinct irreducible quadratic factors

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Repeated quadratic factors

A1, A2, … & B1, B2, … are unique constants A, B, C, D, …

<p>A<sub>1</sub>, A<sub>2</sub>, … &amp; B<sub>1</sub>, B<sub>2</sub>, … are unique constants A, B, C, D, …</p>
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Integral of 1 / (x2 + a2)

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Irreducible quadratic test

If discriminant < 0, then irreducible.

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Solving partial fraction constants

Simultaneous equations OR equating coefficients.

If simulatneous not possible make a variable equal to i = sqrt(-1).

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xn factor

xn is a repeating linear factor: (x - 0)n

<p>x<sup>n</sup> is a <strong>repeating</strong> <strong>linear</strong> factor: (x - 0)<sup>n</sup></p>