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Thread: Cubic Equation

  1. #1
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    Cubic Equation

    I know cubic equations must have the form $\displaystyle Ax^3 + Bx^2 + Cx + D = 0,$ and I know how to solve that type of equation by doing long division etc:


    However I got this cubic question,


    $\displaystyle x^3-3x^2+4$


    It doesn't have the same form as above and it's put me right off knowing what to do. I tried to simplify it $\displaystyle ( x(x^2+3x) +4)$ but I got no where fast. So any starters of where I should start to factorize would be very helpful!
    Last edited by Kaynight; Oct 22nd 2008 at 09:34 PM.
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  2. #2
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    Quote Originally Posted by Kaynight View Post
    I know cubic equations must have the form $\displaystyle Ax^3 + Bx^2 + Cx + D = 0,$ and I know how to solve that type of equation by doing long division etc:


    However I got this cubic question,


    $\displaystyle x^3-3x^2+4$


    It doesn't have the same form as above and it's put me right off knowing what to do. I tried to simplify it $\displaystyle ( x(x^2+3x) +4)$ but I got no where fast. So any starters of where I should start to factorize would be very helpful!
    It has the same form as above, with $\displaystyle C=0$.
    You can notice that $\displaystyle -1$ is a solution, hence $\displaystyle x+1$ divides your polynomial. It remains to compute $\displaystyle Q(x)$ such that $\displaystyle x^3-3x^2+4=(x+1)Q(x)$.
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  3. #3
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    Quote Originally Posted by Laurent View Post
    It has the same form as above, with $\displaystyle C=0$.
    You can notice that $\displaystyle -1$ is a solution, hence $\displaystyle x+1$ divides your polynomial. It remains to compute $\displaystyle Q(x)$ such that $\displaystyle x^3-3x^2+4=(x+1)Q(x)$.
    Many Thanks!
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