# Thread: Eigenvalues

1. ## 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.

2. Yes, factor the cubic you have.

3. Thats great thanks. Can you show me how to do this please?

4. I will use x instead of lambda.

$\displaystyle x^{3}-10x^{2}+28x-24$

Rewrite or use the Rational Root theorem.

Let's rewrite, group, then factor:

$\displaystyle x^{3}-6x^{2}-4x^{2}+24x+4x-24$

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

$\displaystyle x(x^{2}-4x+4)$

Now, factor the remaining ones:

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

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

Put them together:

$\displaystyle x(x^{2}-4x+4)-6(x^{2}-4x+4)$

$\displaystyle (x-6)(x^{2}-4x+4)$

$\displaystyle (x-6)(x-2)^{2}$