Solve for all the roots. Help!!?

Question

x^4 + x^3 + x^2 + x + 1 = 0
The roots are complex. I know that there is a formula for this but after going through my notes I don't think I wrote it down. Please Help!!

Answer 1

x^5 - 1 can be factorised as
x^5 - 1 = (x - 1)(x^4 + x³ + x² + x + 1)

so multiply both sides of
x^4 + x³ + x² + x + 1 = 0
by
x - 1 (provided that x doesn't equal 1)
and you'll have
(x - 1)(x^4 + x³ + x² + x + 1) = 0
i.e.
x^5 - 1 = 0
i.e.
x^5 = 1

So all you have to do now is to find all the solutions to x^5 = 1 (you''l get one real solution (x=1) and 4 complex ones -> discard x=1 and the other 4 solutions are the solutions to the original equation).

[You might also have heard that solving this equation is called looking for 5th roots of unity]

The way to solve x^5 = 1 is by using complex numbers in polar form.
Say x = a + ib (a complex number).
Then it can also be written as
x = |x| · (cos&theta + i · sin&theta)
and
x^5 = |x|^5 · (cos5&theta + i · sin5&theta)

so if x^5 = 1, then |x|^5 = 1 (obviously) and
(cos5&theta + i · sin5&theta) = 1
as there is no term with i on the RHS, it means that
i · sin5&theta = 0
and therefore
cos5&theta = 1
=> 5&theta = 0 + 2k&pi {k = 0, 1, 2, 3, 4}
=> &theta = 2k&pi/5 {k = 0, 1, 2, 3, 4}

So x = |x| · (cos&theta + i · sin&theta) and |x|=1.
For k=0,
x = 1 · (cos0 + i · sin0) = 1 (this is the solution we discard)

And we keep the following four solutions:
For k=1, x = 1 · (cos2&pi/5 + i · sin2&pi/5)
For k=2, x = 1 · (cos4&pi/5 + i · sin4&pi/5)
For k=3, x = 1 · (cos6&pi/5 + i · sin6&pi/5)
For k=4, x = 1 · (cos8&pi/5 + i · sin8&pi/5)

Hope this helps.

Answer 2

Good luck because this equation has no rational roots; there is a formula for this but it is very complex.

Answer 3

Just in case you were wondering why no one ever ever uses the quartic formula, this is why:

http://planetmath.org/encyclopedia/QuarticFormula.html

Its ridiculously long. Interesting there are no such formulae for higher order polynomials, indeed their lack of existence has been proven.

First answerer gives a good method.

Answer 4

This is the text output of wxMaxima. The %i indicated "i", or an imaginary value.

x=-2^(-3/2)*5^(1/4)*
sqrt(sqrt(5)-1)*%i
-sqrt(5)/4-1/4
x=2^(-3/2)*5^(1/4)*
sqrt(sqrt(5)-1)*%i
-sqrt(5)/4-1/4
x=-2^(-3/2)*5^(1/4)*
sqrt(sqrt(5)+1)*%i
+sqrt(5)/4-1/4
x=2^(-3/2)*5^(1/4)*
sqrt(sqrt(5)+1)*%i
+sqrt(5)/4-1/4