a/(ax + c) - b/(bx + d) = (ad - bc)/[(ax + c)(bx + d)]. This one identity suffices to underpin the entire theory of partial fractions. Note that a(-d/b) + c = (-1/b)(ad - bc) b(-c/a) + d = (+1/a)(ad - bc) These truths are neither accidental nor insignificant.
It's easy to see that a(bx + d) - b(ax + c) = ad - bc. Dividing both sides by [(ax + c)(bx + d)] then gives a/(ax + c) - b/(bx + d) = (ad - bc)/[(ax + c)(bx + d)].
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Also note that (ad - bc)/[(a(-d/b) + c)(bx + d)] + (ad - bc)/[(ax + c)(b(-c/a) + d)] = a/(ax + c) - b/(bx + d) = (ad - bc)/[(ax + c)(bx + d)]. This elegant pattern persists throughout the theory of partial fractions, rendering it memorable, and its application simple.
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Oh, wait -- strictly speaking, the theory of partial fractions also requires this identity: (ax + b)/(cx + d) = a/c - (ad - bc)/[c(cx + d)], which comes into play when a the rational function to be decomposed has a numerator (with degree less than that of the denominator.)
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Also, perhaps I ought to have noted that 1/[(ax + c)(bx + d)] = 1/[(a(-d/b) + c)(bx + d)] + 1/[(ax + c)(b(-c/a) + d)]. From the above it then follows that (mx + n)/[(ax + c)(bx + d)] = (m(-d/b) + n)/[(a(-d/b) + c)(bx + d)] + (m(-c/a) + n)/[(ax + c)(b(-c/a) + d)]
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So here's the general story: P(x)/Q(x) = [(mx + n) ... (px + q)]/[(ax + c) ... (bx + d)] = [(m(-d/b) + n) ... (p(-d/b) + q)]/[(a(-d/b) + c) ... (bx + d)] + ... [(m(-c/a) + n) ... (p(-c/a) + q)]/[(ax + c) ... (b(-c/a) + d)], provided deg(P) < deg(Q).
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End of conversation
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