# Thread: Factoring an odd polynomial

1. ## Factoring an odd polynomial

I'm in a Systems & Signals class and I need to find the stability of a system.

I know that if the roots to the characteristic equation of the system are equal or less than 1, the system is stable. My problem is it's been a while since I've had an Algebra course and I haven't had to factor a polynomial like this in a long time.

The equation:

a^4 - (2/3)a - 1/3

How would I go about finding the roots to that polynomial?

2. Originally Posted by Katzenjammer
I'm in a Systems & Signals class and I need to find the stability of a system.

I know that if the roots to the characteristic equation of the system are equal or less than 1, the system is stable. My problem is it's been a while since I've had an Algebra course and I haven't had to factor a polynomial like this in a long time.

The equation:

a^4 - (2/3)a - 1/3

How would I go about finding the roots to that polynomial?
Using the remainder and factor theorems, for any polynomial function $f(x)$

If $f(x_1) = 0$ (where $x_1$ is a number), then $(x - x_1)$ is a factor.

$f(a) = a^4 - \frac{2}{3}a - \frac{1}{3}$

By inspection, we can see that $f(1) = 0$. So $(a - 1)$ is a factor.

Using long division, we can see

$a^4 - \frac{2}{3}a - \frac{1}{3} = (a - 1)\left(a^3 + a^2 + a + \frac{1}{3}\right)$.

I don't believe that this can be factorised further.