# Factoring an odd polynomial

• Feb 2nd 2010, 10:39 PM
Katzenjammer
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?
• Feb 2nd 2010, 10:54 PM
Prove It
Quote:

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 $\displaystyle f(x)$

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

$\displaystyle f(a) = a^4 - \frac{2}{3}a - \frac{1}{3}$
By inspection, we can see that $\displaystyle f(1) = 0$. So $\displaystyle (a - 1)$ is a factor.
$\displaystyle a^4 - \frac{2}{3}a - \frac{1}{3} = (a - 1)\left(a^3 + a^2 + a + \frac{1}{3}\right)$.