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 June 12th, 2013, 09:57 AM #1 Newbie   Joined: Jun 2013 Posts: 1 Thanks: 0 Partial Fraction help! I hate Maths Just to clarify my position on the subject. down to business! I have this partial fraction, which I am told is different from ones I am used to, as Instead of using A and B, we also use C to solve this question! how good. here is problem 3x^3-2x^2-16x+20 ---------------------- (x-2)(x+2)
 June 12th, 2013, 10:50 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Partial Fraction help! You may have misunderstood. You do not need a third constant on a fraction, but since the numerator has degree 3 while the denominator has degree 2, you should first divide to get a "mixed fraction" rather than an "improper fraction". $\frac{x^3- 2x^2- 16x+ 20}{(x- 2)(x+ 2)}= x- 2- \frac{12x- 12}{(x- 2)(x+ 2)}$ Now there exist constants A and B such that $\frac{12x- 12}{(x- 2)(x+ 2)}= \frac{A}{x- 2}+ \frac{B}{x+ 2}$.
 June 12th, 2013, 11:53 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,943 Thanks: 1132 Math Focus: Elementary mathematics and beyond Re: Partial Fraction help! $\frac{3x^3\,-\,2x^2\,-\,16x\,+\,20}{x^2\,-\,4}\,=\,3x\,-\,2\,+\,\frac{12\,-\,4x}{x^2\,-\,4}\,=\,3x\,-\,2\,-\,\frac{5}{x\,+\,2}\,+\,\frac{1}{x\,-\,2}$
 June 12th, 2013, 08:48 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,746 Thanks: 2133 True, but you started with an extra "3" in the numerator.
 June 13th, 2013, 12:38 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,943 Thanks: 1132 Math Focus: Elementary mathematics and beyond Re: Partial Fraction help! To what are you referring?
 June 13th, 2013, 08:05 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,746 Thanks: 2133 Sorry, I should have been saying that [color=#00AA00]HallsofIvy[/color] omitted the leading "3" from the numerator of the fraction.

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