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Math Help - solving for x intercept

  1. #1
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    solving for x intercept

    y=2-15x+9x^2-x^3
    so 0=2-15x+9x^2-x^3
    I don't see anyway to factor this out, does this mean that there is no x intercept?
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  2. #2
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    By graphing it, or otherwise, you can figure out that x=2 is a root. So factor out (x-2) to get (x-2)(x^2-7x+1)=0. Now can you factor the quadratic?
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  3. #3
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    Quote Originally Posted by dorkymichelle View Post
    y=2-15x+9x^2-x^3
    so 0=2-15x+9x^2-x^3
    I don't see anyway to factor this out, does this mean that there is no x intercept?
    all cubic polynomials have at least one real zero.

    using the rational root theorem reveals that x = 2 is a real root of this cubic.

    there may be more ... what can you do now to find them if they exist?
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  4. #4
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    yes thanks, I can find the roots with the quadratic formula, I didnt think of factoring out the (x-2)
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