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Trouble finding x intercepts

  1. Jun 11, 2007 #1
    I'm having trouble finding x intercepts i have the question x^3-9x^2+15x+30 and by setting y=0 i got -30=x(x^2-9x+15) I used the quadratic formula to find that x^2-9x+15 give me no real roots, so the only intercept would be x=0. But this would also mean that the yintercept is 0 and i found that it is 30. What am i doing wrong is it something to do with moving the 30 to the left side?
     
  2. jcsd
  3. Jun 11, 2007 #2

    malawi_glenn

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    -30=x(x^2-9x+15)

    You can not solve the right hand side to be = 0 if you have left hand side = -30 ...
     
  4. Jun 11, 2007 #3
    so how do i do it. i was moving that over to solve for x
     
  5. Jun 11, 2007 #4

    malawi_glenn

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    test solution, then rewrite x^3-9x^2+15x+30 as (x+n)(x-b)(x-4) or something in that style.

    If c solves the p(x) = then p(x) can be divided with (x-c)..
    and vice versa, (x+r) ; P(-r) = 0
     
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