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Math Help - Solving cubic polynomials

  1. #1
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    Solving cubic polynomials

    How do you solve -3x^3 + 7x^2 -14x + 8 = 0?

    It's a characteristic polynomial. Is it possible to solve it from here or must I solve them without opening too many brackets?
    Last edited by geft; October 25th 2009 at 10:14 PM.
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    Quote Originally Posted by geft View Post
    How do you solve -3x^3 + 7x^2 -14x + 8 = 0?

    It's a characteristic polynomial. Is it possible to solve it from here or must I solve them without opening too many brackets?
    The LHS of the equation can't be factored. Therefore you can use either the Cardanic formula (see here: Cubic function - Wikipedia, the free encyclopedia) or an iterative method like Newton's method to get an approximate solution.
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    It shouldn't be that complicated since I wasn't taught that method. Here's the original equation.

     det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0
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    Quote Originally Posted by geft View Post
    How do you solve -3x^3 + 7x^2 -14x + 8 = 0?

    It's a characteristic polynomial. Is it possible to solve it from here or must I solve them without opening too many brackets?
    You could try looking for "rational roots". The "rational root theorem" says that any rational roots of ax^n+ \cdot\cdot\dot+ bx+ c= 0 are of the form x= r/s where s is an integer factor of a, the leading coefficient, and s is an integer factor of c, the constant term.

    Here, the leading coefficient is "3" which has factors 1, -1, 3, and -3. The constant term is "8" which has factors 1, -1, 2, -2, 4, -4, 8, and -8. The only possible rational roots are 1, -1, 2, -2, 4, -4, 8, -8, 1/3, -1/3,, 2/3, -2/3, 4/3, -4/3, 8/3, and -8/3. Trying each of those will either give you a root or show that this equation has no rational roots. If there are no rational roots there will not be any solution simpler than Cardano's.
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    Quote Originally Posted by geft View Post
    It shouldn't be that complicated since I wasn't taught that method. Here's the original equation.

     det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0
    Perhaps you should post the original matrix since it's possible your determinant and hence characteristic equation is wrong ....
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  6. #6
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    You're right. Here's the matrix: <br />
\left(<br />
\begin{array}{ccc}<br />
3 & 1 & -1 \\<br />
-4 & 2 & 2 \\<br />
-2 & 2 & 2<br />
\end{array}<br />
\right)
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    Quote Originally Posted by geft View Post
    It shouldn't be that complicated since I wasn't taught that method. Here's the original equation.

     det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0
    Quote Originally Posted by geft View Post
    You're right. Here's the matrix: <br />
\left(<br />
\begin{array}{ccc}<br />
3 & 1 & -1 \\<br />
-4 & 2 & 2 \\<br />
-2 & 2 & 2<br />
\end{array}<br />
\right)
    You've made a couple or errors in your calculation. It should be:

    det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) {\color{red}- 1}] - 2[{\color{red}-2} - \lambda] = 0.

    Go back and check your work.
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  8. #8
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    <br />
\left(<br />
\begin{array}{ccc}<br />
3 - \lambda & 1 & -1 \\<br />
-4 & 2 - \lambda & 2 \\<br />
-2 & 2 & 2 - \lambda<br />
\end{array}<br />
\right)

    (3 - \lambda) \left| <br />
\begin{array}{cc} 2 - \lambda & 2 \\ <br />
2 & 2 - \lambda<br />
\end{array}<br />
\right| + 4 \left|<br />
\begin{array}{cc} 1 & -1 \\ <br />
2 & 2 - \lambda<br />
\end{array}<br />
\right| - 2 \left|<br />
\begin{array}{cc} 1 & -1 \\ <br />
2 - \lambda & 2<br />
\end{array}<br />
\right|

    det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0

    ???

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    Quote Originally Posted by geft View Post
    <br />
\left(<br />
\begin{array}{ccc}<br />
3 - \lambda & 1 & -1 \\<br />
-4 & 2 - \lambda & 2 \\<br />
-2 & 2 & 2 - \lambda<br />
\end{array}<br />
\right)

    (3 - \lambda) \left| <br />
\begin{array}{cc} 2 - \lambda & 2 \\ <br />
2 & 2 - \lambda<br />
\end{array}<br />
\right| + 4 \left|<br />
\begin{array}{cc} 1 & -1 \\ <br />
2 & 2 - \lambda<br />
\end{array}<br />
\right| - 2 \left|<br />
\begin{array}{cc} 1 & -1 \\ <br />
2 - \lambda & 2<br />
\end{array}<br />
\right|

    det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0

    ???

    OK, I expanded along the top row. Our expresions are equivalent. And give the correct eigenvalues. So your mistake will be somewhere in the simplifying of the expression to a cubic equation. That's where you need to check.
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  10. #10
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    Wow, somehow I got it.

    (3-x)((2-x)(2-x)-4)+4((2-x)+2)-2(4-x)
    (3-x)(4-4x+x^2-4)+4(4-x)-2(4-x)
    (3-x)(-4x+x^2)+(4-2)(4-x)
    (3-x)(-x)(4-x)+2(4-x)
    (-3x+x^2)(4-x)+2(4-x)
    (x^2-3x+2)(4-x)
    (x-2)(x-1)(4-x)=0
    Last edited by geft; October 27th 2009 at 09:12 AM.
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