# Homework Help: Factoring a 3rd order polynomial

1. Dec 23, 2011

### rowardHoark

Factoring a 4th order polynomial

1. The problem statement, all variables and given/known data

Example:
$(jw)^{3}+6(jw)^{2}+5jw+30=0$ can be re-written into $6(5-w^{2})+jw(5-w^{2})$. The fact that there are two identical $(5-w^{2})$ is a desirable outcome. Imaginary number $j=\sqrt{-1}$ becomes -1 when raised to the power of 2.

2. Relevant equations

The problem is to transform $(jw)^{4}+7(jw)^{3}+59(jw)^2+98(jw)+630=0$ in a similar manner.

3. The attempt at a solution

So far I have been unsuccessful.

$w^{4}-7jw^{3}-59w^{2}+98jw+630=0$

$(w^{4}-59w^{2})+7(-jw^{3}+14jw+90)=0$

Last edited: Dec 23, 2011
2. Dec 23, 2011

### CEL

Re: Factoring a 4th order polynomial

Try

$(w^4 - 59 w^2 + 630) + jw(14 - w^2)$

Then divide $(w^4 - 59 w^2 + 630)$ by $(14 - w^2)$

3. Dec 23, 2011

### rowardHoark

Re: Factoring a 4th order polynomial

Thank you, CEL.

The answer is $-(14-w^{2})(w^{2}-45)+7jw(14-w^{2})=0$

If designing a controller using Ziegler-Nichols second method, would I pick $\omega=\sqrt{14}$ or $\omega=\sqrt{45}$ as my value to calculare $P_{cr}=\frac{2\Pi}{\omega}$?

4. Dec 23, 2011

### The Electrician

Re: Factoring a 4th order polynomial

There's one tiny error here:

$(w^4 - 59 w^2 + 630) + jw(14 - w^2)$

should be:

$(w^4 - 59 w^2 + 630) + 7jw(14 - w^2)$

A good systematic method for problems like this is shown in the attachment.

File size:
5 KB
Views:
131