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

[bob1182006]

Factor a 4th order polynomial (Solved)

Homework Statement

Find the roots of:
x^5-1=0

Homework Equations

Polynomial long division.

The Attempt at a Solution

x^5-1 = (x-1)(x^4+x^3+x^2+x+1) = 0
x^4+x^3+x^2+x+1 = (x^2+1)^2+x^3+x-x^2
(x^2+1)^2+x^3+x-x^2 = (x^2+1)^2+x(x^2+1)-x^2

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.

 

[Mark44]

Homework Statement

Find the roots of:
x^5-1=0

Homework Equations

Polynomial long division.

The Attempt at a Solution

x^5-1 = (x-1)(x^4+x^3+x^2+x+1) = 0
x^4+x^3+x^2+x+1 = (x^2+1)^2+x^3+x-x^2
(x^2+1)^2+x^3+x-x^2 = (x^2+1)^2+x(x^2+1)-x^2

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.

 

[Mentallic]

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

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

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

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))

Now if you want only real factors, notice that

(x-cis(\theta))(x-cis(-\theta))=x^2-x(cis(\theta)+cis(-\theta))+cis(\theta)cis(-\theta))

Can you simplify this to get rid of any imaginary numbers?

 

[bob1182006]

Ah thanks, I didn't even think about finding the roots that way.

I just though I could factor out something like (x^2+1)^2 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!

 

[Mentallic]

Yeah I see why you would try that, but basically you wouldn't be able to take a factor of (x^2+1) out since that means there would be roots of \pm i which just isn't the case.