ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Timtasticle on November 10, 2007, 04:18:32 pm
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(http://img246.imageshack.us/img246/6778/imaspudxq0.jpg)
I always stuff these up. Can someone explain how to get the left hand side into the form of the right hand side?
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Long division. :D
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I see, care to elaborate?
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(http://i148.photobucket.com/albums/s4/Toothpick00/longdiv.gif)
hence:
(2x+1)/(x-5)
= 2 + [11 / (x-5)]
hope that made sense.
"answer" plus [ "remainder" / "what you are dividing it by(denominator)" ]
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Oooohh I see. :o
I always try and separate the fractions :oops:
Thanks heaps for your help!!
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No problem, anytime.
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(http://img246.imageshack.us/img246/6778/imaspudxq0.jpg)
I always stuff these up. Can someone explain how to get the left hand side into the form of the right hand side?
Faster way:
(2x+1) / (x-5) = [ 2(x-5) + 10 + 1 ] / (x-5) = 2 + 11/(x-5)
The trick is to force (x-5), the denominator to appear on the numerator, and create a compensation term (like you do with completing the square). I introduced 2(x-5) because we have 2x, so we need the 2 outside, then compensated by adding 10 (because 2*-5 from 2(x-5) would be -10). Added to the original 1, gives 11 as a remainder.
Feel free to ask for clarification, because I did rush that explanation.
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Ah. Good old coblin.
Always to the rescue. :P