# Homework Help: Linear Algebra: Characteristic Equation Factoring

1. Jun 8, 2015

### henry3369

1. The problem statement, all variables and given/known data
If I have the characteristic equation:
3 + 3λ2 + 9λ + 5
And I'm told that one of its eigenvalues is -1.
How do I find the rest of the eigenvalues?

2. Relevant equations
3 + 3λ2 + 9λ + 5

3. The attempt at a solution
The furthest I can get is:
3 + 3λ2 + 9λ + 5 = (λ + 1) x (some factor of the original characteristic equation)

I can't seem to figure out what "some factor of the original characteristic equation" is. I tried it by inspection and I don't see how (λ + 1) helps. Is there some method that I'm missing to find that part?

2. Jun 8, 2015

### ehild

Apply polynomial long division http://en.wikipedia.org/wiki/Polynomial_long_division.

3. Jun 8, 2015

### MarcusAgrippa

First, you have not written an equation, but a cubic expression. Probably an oversight on your part.

Your expression contains $- \lambda^3$. So it must arise from an expression of the form $- (\lambda -a)(\lambda -b)(\lambda -c)$. You can get rid of the negative in the characteristic equation by simple algebra. You say correctly that one factor of the remaining cubic must be $(\lambda - (-1) )$ or $(\lambda + 1)$. A simple long division of the given expression (with the signs changed) will leave you with a quadratic expression. You can then use the standard quadratic formula to factorise the quadratic.

4. Jun 8, 2015

### Zondrina

It's not always the case a cubic polynomial will factor into that form. In this problem you have three linear factors, which is nice. In general though, there could be an irreducible quadratic generated by the long division. A more general form would be:

$$ax^3 + bx^2 + cx + d = (x \pm e)(fx^2 \pm gx \pm h^2)$$

Where the cubic is factored into the product of a linear factor and a quadratic factor. The quadratic factor may or may not be reducible depending on where you are.

Also see: http://mathworld.wolfram.com/CubicFormula.html

Last edited: Jun 8, 2015
5. Jun 8, 2015

### ehild

The characteristic equation can have complex roots: the eigenvalues can be complex. And with them, it factors into three linear factors.

Last edited: Jun 8, 2015
6. Jun 9, 2015

### Zondrina

I completely ignored $\mathbb{C}$ there. My bad.

7. Jun 9, 2015

### HallsofIvy

While it is true that any cubic polynomial can be factored to three linear terms, I think I would first start looking for a quadratic factor- call it $a\lambda^2+ b\lambda+ c$. Then we must have $(\lambda+ 1)(a\lambda^2+ b\lambda+ c)= (a\lambda^3+ b\lambda^2+ c\lambda)+ (a\lambda^2+ b\lambda+ c)= a\lambda^3+ (a+ b)\lambda^2+ (b+ c)\lambda+ c= -\lambda^3+ 3\lambda^2+ 9\lambda+ 5$
so that we must have $a= -1$, $a+ b= 3$, $b+ c= 9$, and $c= 5$

That is, of course, 4 equations for 3 unknowns. It is because we know that x+ 1 is a factor that we known these equations are consistent.