Factoring a 3rd order polynomial

In summary, the problem is to transform the equation (jw)^{4}+7(jw)^{3}+59(jw)^2+98(jw)+630=0 in a similar manner to the given example. After unsuccessful attempts, a possible solution is found by grouping the terms and dividing by (14 - w^2). However, there is a small error in the initial attempt that should be corrected. A good systematic method for solving such problems is shown in the attached document.
  • #1
rowardHoark
15
0
Factoring a 4th order polynomial

Homework Statement



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

Homework Equations



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

The Attempt at a Solution



So far I have been unsuccessful.

[itex]w^{4}-7jw^{3}-59w^{2}+98jw+630=0[/itex]

[itex](w^{4}-59w^{2})+7(-jw^{3}+14jw+90)=0[/itex]
 
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  • #2


rowardHoark said:

Homework Statement



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

Homework Equations



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

The Attempt at a Solution



So far I have been unsuccessful.

[itex]w^{4}-7jw^{3}-59w^{2}+98jw+630=0[/itex]

[itex](w^{4}-59w^{2})+7(-jw^{3}+14jw+90)=0[/itex]

Try

[itex](w^4 - 59 w^2 + 630) + jw(14 - w^2)[/itex]

Then divide [itex](w^4 - 59 w^2 + 630)[/itex] by [itex](14 - w^2)[/itex]
 
  • #3


CEL said:
Try

[itex](w^4 - 59 w^2 + 630) + jw(14 - w^2)[/itex]

Then divide [itex](w^4 - 59 w^2 + 630)[/itex] by [itex](14 - w^2)[/itex]

Thank you, CEL.

The answer is [itex]-(14-w^{2})(w^{2}-45)+7jw(14-w^{2})=0[/itex]

If designing a controller using Ziegler-Nichols second method, would I pick [itex]\omega=\sqrt{14}[/itex] or [itex]\omega=\sqrt{45}[/itex] as my value to calculare [itex]P_{cr}=\frac{2\Pi}{\omega} [/itex]?
 
  • #4


CEL said:
Try

[itex](w^4 - 59 w^2 + 630) + jw(14 - w^2)[/itex]

Then divide [itex](w^4 - 59 w^2 + 630)[/itex] by [itex](14 - w^2)[/itex]

There's one tiny error here:

[itex](w^4 - 59 w^2 + 630) + jw(14 - w^2)[/itex]

should be:

[itex](w^4 - 59 w^2 + 630) + 7jw(14 - w^2)[/itex]

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

Attachments

  • Nichols.png
    Nichols.png
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  • #5


(w^{2}-59)(w^{2}+14j)+7j(w^{2}+14j)=0

At this point, I am unable to simplify the expression any further. It is important to note that factoring a 3rd order polynomial and a 4th order polynomial are two different processes. In fact, factoring a 4th order polynomial is more complex and often involves using techniques such as grouping, the rational root theorem, and the quadratic formula. In this case, the polynomial is not easily factorable and may require the use of a computer program or graphing calculator to find the roots. It is also possible that the polynomial may not have any real roots and will need to be factored using complex numbers. As a scientist, it is important to carefully analyze and approach each problem with the appropriate techniques and tools.
 

Related to Factoring a 3rd order polynomial

1. What is a 3rd order polynomial?

A 3rd order polynomial is a polynomial expression of the form ax^3 + bx^2 + cx + d, where a, b, c, and d are constants and x is a variable. It is also known as a cubic polynomial.

2. Why is factoring a 3rd order polynomial important?

Factoring a 3rd order polynomial is important because it helps us find the roots or solutions of the polynomial. This can be useful in solving equations and understanding the behavior of the polynomial function.

3. How do you factor a 3rd order polynomial?

To factor a 3rd order polynomial, we can use various methods such as grouping, difference of cubes, and sum of cubes. The goal is to rewrite the polynomial as a product of simpler polynomials or binomials.

4. Can all 3rd order polynomials be factored?

No, not all 3rd order polynomials can be factored. Some polynomials may have complex roots or irrational roots, which cannot be factored using real numbers. However, the rational root theorem can help us determine if a polynomial can be factored using integer roots.

5. What are the applications of factoring a 3rd order polynomial?

Factoring a 3rd order polynomial has various applications in fields such as physics, engineering, and economics. It can help us solve real-world problems, model and analyze data, and understand the behavior of complex systems.

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