Yes, factor the cubic you have.
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.
I will use x instead of lambda.
Rewrite or use the Rational Root theorem.
Let's rewrite, group, then factor:
Using , factor out an x:
Now, factor the remaining ones:
See?. What's in the parentheses is the same. That is imperative when factoring.
Put them together: