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Math Help - Eigenvalues

  1. #1
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    Eigenvalues

    I am working on eigenvalues, I have got my equation down to
    λ^3 10λ^2 + 28λ - 24

    I know that λ (eigenvalues) are equal to 2, 2 and 6 but this is because the example was done by getting the solution (λ-2)^2 (-λ+6) = 0

    Is there a way I can get that answer by using λ^3 10^λ2 + 28λ - 24.
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  2. #2
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    Yes, factor the cubic you have.
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  3. #3
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    Thats great thanks. Can you show me how to do this please?
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  4. #4
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    I will use x instead of lambda.

    x^{3}-10x^{2}+28x-24

    Rewrite or use the Rational Root theorem.

    Let's rewrite, group, then factor:

    x^{3}-6x^{2}-4x^{2}+24x+4x-24

    Using x^{3}-4x^{2}+4x, factor out an x:

    x(x^{2}-4x+4)

    Now, factor the remaining ones:

    -6x^{2}+24x-24=-6(x^{2}-4x+4)

    See?. What's in the parentheses is the same. That is imperative when factoring.

    Put them together:

    x(x^{2}-4x+4)-6(x^{2}-4x+4)

    (x-6)(x^{2}-4x+4)

    (x-6)(x-2)^{2}
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