Find roots to quintic polynomial

[VeeEight]

Homework Statement

Find the roots to the polynomial x⁴+x+1

this is a problem I am stuck with for my algebra class

Homework Equations

there are a few tricks I have learned for finding roots like taking conjugates and rewriting using fundamental theorem of algebra but we haven't worked on this stuff too much

The Attempt at a Solution

by fundamental theorem of algebra, we can write x⁴+x+1 = (x-a)(x-b)(x-c)(x-d) for some complex numbers a,b,c,d

so the roots a,b,c,d have to follow the equations:
a + b + c + d = 0
ad + ab + bd + cd + bc = 0
abd + bdc + acd + abc = 1
abcd = 1

other than that I do not know what else to do to find the roots

 

[Mark44]

Any complex numbers that are solutions show up in pairs as conjugates a + bi and a - bi.

Also, this is a quartic (degree 4), not a quintic (degree 5).

 

[VeeEight]

Any complex numbers that are solutions show up in pairs as conjugates a + bi and a - bi.

Also, this is a quartic (degree 4), not a quintic (degree 5).

Yup the part about conjugates is one of the only things I know about this subject
and thanks, I put down quintic when I should have said quartic.

 

[lanedance]

have a look at your equation, can you convince yourself there are no real roots? (ie is f(x) > 0 for all x in the reals?

if so as Mark implied, you know there will be 2 sets of complex conjugate roots, try inputting these in your equation and solving

 

[VeeEight]

okay so the roots are of the form: r₁ = a+ib, r₂ = a-ib, r₃ = c+id, r₄ = c-id where a,b,c,d are reals.

then we obtain the equations 2a + 2c = 0 and
(a²+b²)(c²+d²) = 1

 

[lanedance]

on the right track, but you should get an equation for each coefficient of x

 

[VeeEight]

okay so the equations are:
2a + 2c = 0
(a² + b²) (c² + d²) = 1
2(a²c + ac² + ad²+b²c) + i (a²b + d²b + c²b+b²d + a²d + c²d + d³ + b³) = 0
a² + b² + c² + d² + 3ac + bc -i (ad + ac) = 1

 

[lanedance]

multiple the conjugate pairs first

for a complex number z = a + ib
z.z* = (a^2 + b^2) is real
z+z* = 2a is real
so (x-z)(x-z*) = x^2 - (z+z*)x + z.z* is then real

so there should be no coefficents of i in your equations, and there should be no d³ as it only appears in factors twice...

 

[VeeEight]

I always seem to get coefficients of i in the equations though

 

[lanedance]

try this
for
z = a + ib, z* = a-ib
y = c + id, y* = c-id

(x-z)(x-z*)(x-y)(x-y*) = (x^2 - (z+z*)x + z.z*)(x^2 - (y+y*)x + y.y*)

there is clearly no i's left on the right hand side