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Math help...please?

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What's the answer?
  A) { 2, 1, -7/3 }
  B) { 2, -1, 7/3 }
  C) { 2, 7, -1/3 }
  D) { 2, -7, 1/3 }
  pfft, I don't know. (poll whore)
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Keakealani

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PostPosted: Sun May 27, 2007 4:32 pm


Okay, first off, here's the problem.
User Image - Blocked by "Display Image" Settings. Click to show.
Sorry for the image-ness, but I can't do exponents in forum code, I'm pretty sure.

Anyway, I don't need help as much with the problem (although of course that would be nice) as much as the concept. I read the chapter of my textbook and relevant Wikipedia articles, but I still don't really understand how you solve based on the zero that's given.

According to my answer key, the correct answer is (highlight)
C) { 2, 7, -1/3 }

~ edit ~
I got help from one of my friends, so I guess never mind.

For anyone who might be interested, the answer is that if you know 2 is a zero then (x-2) is a factor, and then you can divide the polynomial by (x-2) and come up with a second-degree polynomial which should be factorable - the end result is (x-2)(x-7)(3x+3)=0

and as a note synthetic division does fine here.
PostPosted: Tue Aug 21, 2007 2:30 am


I think you're supposed to factorise the expressions, then substitute x for 0, because when the line is plotted on a graph, the y axis is the same as x = 0. We haven't quite covered how to do this in class, but we've done quadratics, and it appears to be similar.

zingebar

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