Solving an equation in 4th degree

[Abdul Quadeer]

1. The problem statement, all variables and given/known data
How to solve x4 + x3 + x2 + x + 1= 0 ?

[The legend]

Any 4th degree eqn can be solved by Ferrari's method
http://www.proofwiki.org/wiki/Ferrari%27s_Method

by the way ... how did you attempt it?
Did you try the "substitute some random value" method?

[Abdul Quadeer]

Yes, I tried it but it did not work. Ferrari's method looks good.

[The legend]

it IS good! :biggrin: (sure shot solution guaranteed)

[HallsofIvy]

x^4+ x^3+ x^2+ x+ 1 is a "cyclotomic" polynomial whose zeros are equally spaced around the unit circle in the complex plane.

One way to solve this is to convert it to a 5th degree equation (for which there is no general formula!). Multiplying x^4+ x^3+ x^2+ x+ 1= 0 by x- 1 gives (x- 1)(x^4+ x^3+ x^2+ x+ 1)= (x^5+ x^4+ x^3+ x)- (x^4+ x^3+ x+ 1)= x^5- 1= 0.

Now, the roots of x^5- 1= 0 are the complex numbers e^{2\pi ki/5} where k goes from 0 to 4. Obviously, k= 0 gives x= 1 which statisfies x- 1= 0 but not x^4+ x^3+ x^2+ 1= 0 so the roots of x^4+ x^3+ x^2+ x+ 1 are
e^{2\pi ki/5}= cos(\frac{2\pi k}{5})+ i sin(\frac{2\pi k}{5})
for k= 1, 2, 3, and 4.

[awkward]

I like HallsOfIvy's solution, but just for variety here is another approach.

Divide by x^2, yielding

x^2 + x + 1 + x^{-1} + x^{-2} = 0.

Now let y = x + x^{-1}.

With this substitution, the equation becomes

y^2 + y - 1 = 0.

Solve for y (two roots), then solve for x (four roots).

[The legend]

nice one awkward...HallsofIvy has an nice solution too(one i never thought of)...

[Abdul Quadeer]

Divide by LaTeX Code: x^2 , yielding

LaTeX Code: x^2 + x + 1 + x^{-1} + x^{-2} = 0 .

Now let LaTeX Code: y = x + x^{-1} .

With this substitution, the equation becomes

LaTeX Code: y^2 + y - 1 = 0 .

Solve for y (two roots), then solve for x (four roots).

That is a very ingenious solution. Thanks.