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PartialFractions

# PartialFractions - 3 The denominator contains irreducible...

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A quick summary of partial fractions Math 408C: Di/erential and Integral Calculus Unique Numbers 55420, 55425, and 55430 We apply the technique of partial fractions to a rational function F ( x ) = P ( x ) Q ( x ) ; where both P and Q are polynomials. We assume that long division has already been done, leaving us with deg( P ) < deg( Q ) . The next step is to factor the denominator Q ( x ) . There are then four possible cases. 1. The denominator is the product of linear factors, none of which are repeated, say Q ( x ) = ( ax + p )( bx + q )( cx + r )( dx + s ) : In this case, the partial fraction decomposition takes the form F ( x ) = A ax + p + B bx + q + C cx + r + D dx + s ; where A; B; C; : : : are constants that we can solve for. 2. The denominator is the product of linear factors, some of which are repeated, say Q ( x ) = ( ax + p ) 3 ( bx + q )( cx + r )( dx + s ) : In this case, the partial fraction decomposition takes the form F ( x ) = A 1 ax + p + A 2 ( ax + p ) 2 + A 3 ( ax + p ) 3 + B bx + q + C cx + r + D dx + s ; where A 1 ; A 2 ; A 3 and B; C; D : : : are again constants we solve for.
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Unformatted text preview: 3. The denominator contains irreducible quadratic factors, none of which are repeated. (We say a quadratic is irreducible if it cannot be factored using real numbers; for example, x 2 +1 is irreducible.) If, say, Q ( x ) = ( ax 2 + p 2 )( bx + q )( cx + r )( dx + s ) ; with ax 2 + p 2 irreducible, then the partial fraction decomposition will be F ( x ) = A 1 x + A 2 ax 2 + p 2 + B bx + q + C cx + r + D dx + s : 4. The denominator contains a repeated irreducible quadratic factor, say Q ( x ) = ( ax 2 + p 2 ) 3 ( bx + q )( cx + r )( dx + s ) : In this case, the partial fraction decomposition takes the form F ( x ) = A 1 x + A 2 ax 2 + p 2 + A 3 x + A 4 ( ax 2 + p 2 ) 2 + A 5 x + A 6 ( ax 2 + p 2 ) 3 + B bx + q + C cx + r + D dx + s :...
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