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Q&A #6121 |
From: Michael Sakowski
To: Teacher2Teacher Public Discussion
Date: 2007091307:52:49
Subject: Polynomial Fractions
The equation for resistors in parallel comes to mind. See http://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.intro.parallel.html We use the inverse of this operation (partial fraction decomposition) a lot in calculus and differential equations. For example, to find the integral of 1/(x^2 - x - 6) we would "un-add" and break up this expression into two simpler rational expressions using partial fraction decomposition. We would get the integral of [(1/5)/(x-3) + (-1/5)/(x+2)] which could then easily be integrated. In order to check the decomposition of the above result (or even understand the process), one must know how to add rational expressions.
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