Can a 4th Order Polynomial be Factored Without a Computer?

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And no problem, glad to help!In summary, the conversation discussed the process of factoring a 4th order polynomial and finding the roots of a 5th order polynomial. The solution involved using polynomial long division and representing the complex roots in rectangular form. It was also mentioned that finding the real factors can be done by simplifying the expression (x-cis(\theta))(x-cis(-\theta)).
  • #1
bob1182006
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Factor a 4th order polynomial (Solved)

Homework Statement


Find the roots of:
[tex] x^5-1=0[/tex]

Homework Equations


Polynomial long division.

The Attempt at a Solution


[tex] x^5-1 = (x-1)(x^4+x^3+x^2+x+1) = 0[/tex]
[tex] x^4+x^3+x^2+x+1 = (x^2+1)^2+x^3+x-x^2[/tex]
[tex] (x^2+1)^2+x^3+x-x^2 = (x^2+1)^2+x(x^2+1)-x^2[/tex]

Stuck at this point, I just can't seem to factor out something useful.

I know all of the roots are complex but I need to be able to solve the problem without a computer.
 
Last edited:
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  • #2
bob1182006 said:

Homework Statement


Find the roots of:
[tex] x^5-1=0[/tex]

Homework Equations


Polynomial long division.

The Attempt at a Solution


[tex] x^5-1 = (x-1)(x^4+x^3+x^2+x+1) = 0[/tex]
[tex] x^4+x^3+x^2+x+1 = (x^2+1)^2+x^3+x-x^2[/tex]
[tex] (x^2+1)^2+x^3+x-x^2 = (x^2+1)^2+x(x^2+1)-x^2[/tex]

Stuck at this point, I just can't seem to factor out something useful.

I know all of the roots are complex but I need to be able to solve the problem without a computer.
Four of the roots of x5 - 1 = 0 are complex and one is real (x = 1). The complex roots are located around the unit circle at 72 deg, 144 deg, 216 deg, and 288 deg. These can be represented in rectangular form, with the first one being cos(72 deg) + i sin(72 deg). The others can be represented similarly. I don't know if there's going to be a way to factor your fourth-degree factor.
 
  • #3
It depends what you want, do you want that 4th degree factored among the reals or factored into its linear roots?

What you should do is find all the complex roots as Mark has done, so what you have is

[tex]x_1=1[/tex]
[tex]x_2=cis(2\pi/5)[/tex]
[tex]x_3=cis(4\pi/5)[/tex]
[tex]x_4=cis(-2\pi/5)[/tex]
[tex]x_5=cis(-4\pi/5)[/tex]

So since these are all its roots, to factorize it into its linear roots it's simple.

[tex]x^5-1=(x-1)(x-cis(2\pi/5))(x-cis(4\pi/5))(x-cis(-2\pi/5))(x-cis(-4\pi/5))[/tex]

Now if you want only real factors, notice that

[tex](x-cis(\theta))(x-cis(-\theta))=x^2-x(cis(\theta)+cis(-\theta))+cis(\theta)cis(-\theta))[/tex]

Can you simplify this to get rid of any imaginary numbers?
 
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  • #4
Ah thanks, I didn't even think about finding the roots that way.

I just though I could factor out something like [itex](x^2+1)^2[/itex] which would then give me the complex roots.

@Mentallic, the sines cancel out so you're left with just cosines.

Changed the title since the problem's solved.

Thanks a lot!
 
  • #5
Yeah I see why you would try that, but basically you wouldn't be able to take a factor of [tex](x^2+1)[/tex] out since that means there would be roots of [itex]\pm i[/itex] which just isn't the case.
 

1. What is a 4th order polynomial?

A 4th order polynomial, also known as a quartic polynomial, is a function of the form f(x) = ax4 + bx3 + cx2 + dx + e, where a, b, c, d, and e are constants and x is the variable.

2. How do you factor a 4th order polynomial?

To factor a 4th order polynomial, you can use the grouping method or the factoring by grouping method. These methods involve finding common factors between terms and grouping them together to simplify the expression.

3. What is the difference between factoring and solving a 4th order polynomial?

Factoring a 4th order polynomial involves breaking it down into simpler expressions, while solving a 4th order polynomial involves finding the values of x that make the polynomial equal to zero.

4. Can all 4th order polynomials be factored?

No, not all 4th order polynomials can be factored. Some polynomials may have complex or irrational roots that cannot be factored using integers.

5. Why is factoring a 4th order polynomial important in mathematics?

Factoring a 4th order polynomial is important because it allows us to simplify and solve complex expressions, which is useful in various mathematical applications such as finding roots, graphing, and solving equations.

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