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 March 18th, 2017, 06:32 PM #1 Member   Joined: Mar 2017 From: Tasmania Posts: 36 Thanks: 2 Show that the function can be written as... a) Given the function f(x)=(2x^2-12x+15)/〖(x-3)〗^2 ;x≠3, show that the function can be written as f(x)=(-3)/〖(x-3)〗^2 +2;x≠3
 March 18th, 2017, 06:46 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,780 Thanks: 919 $f(x) = \dfrac{2x^2 - 12x + 15}{(x-3)^2},~x \neq 3$ $2(x-3)^2 = 2(x^2 - 6x + 9) = 2x^2 -12x + 18$ $2x^2 - 12x + 15 = 2(x-3)^2 - 3$ $f(x) = \dfrac{ 2(x-3)^2 - 3}{(x-3)^2} =$ $\dfrac{2(x-3)^2}{(x-3)^2}-\dfrac {3}{(x-3)^2} =$ $2 -\dfrac{3}{(x-3)^2}$ Thanks from topsquark and Posher
 March 18th, 2017, 06:50 PM #3 Senior Member   Joined: May 2016 From: USA Posts: 916 Thanks: 366 This is a little tricky. $\dfrac{2x^2 - 12x + 15}{(x - 3)^2} = \dfrac{2x^2 - 12x + 15 + 3 - 3}{(x - 3)^2} = \dfrac{2(x^2 - 6x + 9) - 3}{(x - 3)^2}.$ Now what?
 March 18th, 2017, 07:00 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,757 Thanks: 1008 Math Focus: Elementary mathematics and beyond $$\frac{2x^2-12x+15}{(x-3)^2}=\frac{2x^2-11x+15-x}{(x-3)^2}=\frac{2x-5}{x-3}-\frac{x}{(x-3)^2}$$ $$\frac{2x-6+1}{x-3}=2+\frac{1}{x-3},\,2+\frac{1}{x-3}-\frac{x}{(x-3)^2}=2+\frac{-3}{(x-3)^2}$$
 March 18th, 2017, 07:36 PM #5 Member   Joined: Mar 2017 From: Tasmania Posts: 36 Thanks: 2 @romsek thanks for this I managed to see how to do it legit like 10 minutes after posting your way was way the way I did it but missed out a few steps so thank you for showing me them

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