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PartialFraction Decomposition: Examples (page 3 of 3) Sections: General techniques, How to handle repeated and irreducible factors, Examples If the denominator of your rational expression has repeated unfactorable quadratics, then you use linearfactor numerators and follow the pattern that we used for repeated linear factors in the denominator; that is, you'll use fractions with increasing powers of the repeated factors in the denominator.
Since x^{2} + 1 is not factorable, I'll have to use numerators with linear factors. Then the decomposition setup looks like this: Thankfully, I don't have to try to solve this one. One additional note: Partialfraction decomposition only works for "proper" fractions. That is, if the denominator's degree is not larger than the numerator's degree (so you have, in effect, an "improper" polynomial fraction), then you first have to use long division to get the "mixed number" form of the rational expression. Then decompose the remaining fractional part.
The numerator is of degree 5; the denominator is of degree 3. So first I have to do the long division: The long division rearranges the rational expression to give me: Now I can decompose the fractional part. The denominator factors as (x^{2} + 1)(x – 2). The x^{2} + 1 is irreducible, so the decomposition will be of the form: Multiplying out and solving, I get: 2x^{2} + x
+ 5 = A(x^{2} + 1) + (Bx + C)(x
– 2) Then the complete expansion is: The preferred placement of the "minus" signs, either "inside" the fraction or "in front", may vary from text to text. Just don't leave a "minus" sign hanging loose underneath. << Previous Top  1  2  3  Return to Index



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