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Find roots to quintic polynomial

  • Thread starter VeeEight
  • Start date
1. Homework Statement
Find the roots to the polynomial x4+x+1

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

2. 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


3. 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
 
32,575
4,306
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).
 
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

Homework Helper
3,304
2
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
 
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

Homework Helper
3,304
2
on the right track, but you should get an equation for each coefficient of x
 
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

Homework Helper
3,304
2
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...
 
I always seem to get coefficients of i in the equations though
 

lanedance

Homework Helper
3,304
2
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
 
Last edited:

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