Factoring a 3rd order polynomial

[rowardHoark]

Factoring a 4th order polynomial

Homework Statement

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.

Homework Equations

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

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

 

[CEL]

Homework Statement

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 disirable outcome. Imaginary number j=\sqrt{-1} becomes -1 when raised to the power of 2.

Homework Equations

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

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

Try

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

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

 

[rowardHoark]

Try

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

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

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}?

 

[The Electrician]

Try

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

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

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.