f(x)=x^5-1?

Question

I know that it has a root at x=1

It also obviously has some imaginary roots being it is a 5th degree polynomial...and only has one real root.....but how would I go about finding the imaginary roots?

hahaha....everyone look at jeffrey7000's comment...he's so funny *sighs*

Answer 1

I would transform it to a complex polar form.

As a complex number 1 can be written as 1 + 0i which is also equal to 1cis0º (abbrev. for 1(cos0º+isin0º)

The five fifth roots in complex form are

1cis0º which is of course the same as 1
1cis72º
1cis144º
1cis 216º
1cis288º

The last 4 are the complex roots.

I don't feel any need to translate those back to rectangular form. I hope that makes sense.

Answer 2

First, factor out the 1.

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

So now we need to solve x^4+x^3+x^2+x+1 = 0. That's a messy quartic equation. However, note that the polynomial is a palindrome; the same backwards and forwards. Substitute 1/y for x in a polynomial equation and you get another equation whose coefficients are backwards of the first. So if 2 is a root of x^2-5x+6 = 0, then 1/2 is a root of 6x^2-5x+1 = 0.

Since x^4+x^3+x^2+x+1 is a palindrome, it is its own backward self. Therefore if r is a root, then 1/r is also a root. The two roots satisfy a quadratic equation x^2-(r+1/r)x+1 = 0. Therefore, to factor x^4+x^3+x^2+x+1, we need to break it up into two quadratics

(x^2+Ax+1)(x^2+Bx+1) = 0,

which is the same as saying

x^4+(A+B)x^3+(2+AB)X^2+(A+B)x+1 = 0

Compare this with x^4+x^3+x^2+x+1 = 0. Set the coefficients equal to each other:

1 = 1
A+B = 1
2+AB = 1
A+B = 1
1=1

Since A+B = 1, B = 1-A. So 2+AB = 2+A(1-A)= 1 or

A^2-A-1 = 0

The roots of this equation are the golden ratios phi = (sqrt(5)+1)/2 and phj = (-sqrt(5)+1)/2. So A = phi, say. This means B = 1-A = phj. Therefore, the equation is now:

(x^2+phi*x+1)(x^2+phj*x+1) = 0

This means the roots of the equation are, using the quadratic formula,

(-phi +/- sqrt(phi^2-4))/2 and
(-phj +/- sqrt(phj^2-4))/2

If you expand this into sqrt(5) notation, I believe you get

(+/-sqrt(5)+1)/4 +/- i * (10 +/- 2*sqrt(5))/4

where the choice of +/- signs has to be chosen correctly. If you have a TI-89, this is the answer it will give you.

Answer 3

Factor x^5 -1

(x - 1)(x^4 +x^3 + x^2 + x + 1)

Then the imaginary roots of x^5 -1 = 0

are roots of x^4 +x^3 + x^2 + x + 1 = 0

Divide both sides by x^2

x^2 + 1/x^2 + x + 1/x + 1 = 0

We have x^2 + 1/x^2 = (x + 1/x)^2 - 2

Then by substitution

(x + 1/x)^2 + (x + 1/x) -1 = 0

Solve for x + 1/x using quadratic formula

x + 1/x = [-1 - sqrt(5)]/2

or x + 1/x = [-1 + sqrt(5)]/2

To solve the first

x + 1/x = [-1 - sqrt(5)]/2

Multiply both sides by x

Then solve

x^2 +[1 + sqrt(5)]x/2 + 1 = 0 for x
Use quadratic formula
Similarly to solve

x + 1/x = [-1 + sqrt(5)]/2

for x, multiply both sides by x and arrange
as a quadratic equation and use quadratic
formula

Answer 4

Rewrite x^5 - 1 = 0 as x^5 = 1. Now, write 1 in its polar form, e^0. That's 1*e^0i, where 1 is R, the modulus of the complex number, in this case equal to the absolute value of the real number 1, and 0 is the value of theta, which is zero for a positive real number. i is the imaginary number, the square root of -1. Hopefully you've already learned the polar form of complex numbers. Now, to get x^5 = e^0i, you need to set x = R*e^ti, where (R*e^ti)^5 = e^0i. Expanding this, we have (R^5)*e^5ti = e^0i, which we can separate into R^5 = 1 and 5t = 0. R is easy to find, because R^5 = 1 has only one solution, R = 1. Imaginary roots aren't used when you're looking for the modulus. t is harder. Initially, it appears that 5t = 0 has only the solution t = 0, corresponding to e^0 = 1, the answer we already had. But remember, this is polar. In polar coordinates, 2kπ is the same as 0, where k is any integer. You need to use as many different k as will give you different answers for t. k = 0 already gave us t = 0. k = 1 gives us 5t = 2π and t = 2π/5. k = 2 gives us t = 4π/5. k = 3 gives us t = 6π/5. k = 4 gives us t = 8π/5. k = 5 gives us t = 10π/5 = 2π = 0, which we already had, so we don't need to use that one, and we're done. The four additional roots are e^ti, where the t are given above. If you need to convert the polar numbers back to complex numbers, it's just e^ti = cos(t) + i*sin(t).

Answer 5

Divide x^5-1 by (x-1), and you obtain a quartic equation:

x^4 + x^3 + x^2 + x + 1

You can then use the (nontrivial) techniques given in the sources below (or the quartic root calculator at the last source) to find the remaining roots.

Answer 6

It seems to be in line with an early Tribulation Epistle, which Peter is (1 Pet 1). He's addressing this to the "strangers" and that would go with those Rev 7 144,000. Here Peter is exhorting them through the witness of Christ while He was in the flesh (human flesh - not the body He's in now - you're version has it wrong) that as He suffered and went through it, then you too can go through this time of Tribulation. If you remember that the Books starting with Hebrews is written for those who are going through the Tribulation and the birth pangs of the Millennium then it becomes less mysterious.

Answer 7

f(x)=x^5-1=(x-1){x^4 + x^3 +x^2+ x +1).............i
{x^4 + x^3 +x^2+ x +1)=x^2{x^2+x + 1 +1/x + 1/x^2}
{x^4 + x^3 +x^2+ x +1)=x^2{(x^2+ 1/x^2)+ !+(x +1/x) }.........ii
Put (x + 1/x) = a
squaring both sides we get
(x + 1/x)^2 = a^2
x^2 +2 +1/x^2= a^2
x^2 +1/x^2= a^2 - 2............................iii
From ii and iii
{(x^2+ 1/x^2)+ 1+(x +1/x) }=a^2 - 2 +1+a
{(x^2+ 1/x^2)+ 1+(x +1/x) }=a^2 + a - 1
= {a +(1+sq rt5)/2 } {a +(1-sq rt5)/2 }
putting the value of a we can have
= {(x+i/x)+(1+sq rt5)/2 } {(x+1/x) +(1-sq rt5)/2 }
Each bracket can be saved further
f(x)=x^5-1=(x-1){(x^2+1)+(1+sq rt5)x/2 } {(x^2+1) +(1-sq rt5)x/2 }
like this we can solve further
therefore x= -1 - [Sq(2 times sq rt 5 -10)]............
therefore x= -1 + [Sq(2 times sq rt 5 -10)]..........
Similarly x= -1 - [-Sq(2 times sq rt 5 -10)]--------
x = -1 +[-Sq(2 times sq rt 5 -10)]--------
Therefore five roots are
x= 1
x= -1 - [Sq(2 times sq rt 5 -10)]
x= -1 + [Sq(2 times sq rt 5 -10)]
x= -1 - [-Sq(2 times sq rt 5 -10)
x = -1 +[-Sq(2 times sq rt 5 -10)

Answer 8

I died trying to figure out what "?" meant...