Finding roots of x^4 - x^3 + x^2 - x + 1 = 0

[Ruby]

Hello, I am having an extremely hard time finding the roots of this equation:

x^4 - x^3 + x^2 - x + 1 = 0

This is the work that I have done:

I pulled an x^2 out:

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

So, a root would be positive one, but how can I find the other three?

 

[jwpaine]

There are no real solutions to this polynomial. You can tell right off because the leading coefficient 1 (only factor of 1/1a) when evaluated f(-1) or f(1) doesn't = 0

There are 4 imaginary solutions.

You can't pull an x^2 out because what would happen to your other x and -1?

You are close... but not close enough:

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


Maybe someone else can help you find the imaginary solutions....(I'm not quite sure how to approach it)

I hope this helps you a bit.


Would someone please show Ruby and I how to find the imaginary roots?

 

[soroban]

Re: Finding roots

Hello, Ruby!

Is there a typo in the equation? . . . There are no real roots.

\(\displaystyle x^4\,-\,x^3\,+\,x^2\,-\,x\,+\,1\:=\:0\)

Multiply both sides by \(\displaystyle (x\,+\,1)\)
. . and we get: \(\displaystyle \:x^5\,+\,1\:=\:0\;\;\Rightarrow\;\;x^5 \:=\:-1\)

The solution is the five fifth roots of -1.

One of them is \(\displaystyle x\,=\,-1\) (which we introduced)
. . and the other four are complex numbers,
. . namely: \(\displaystyle \,\text{cis}36^o,\,\text{cis}108^o,\,\text{cis}252^o,\,\text{cis}324^o\)

 

[stapel]

x^4 - x^3 + x^2 - x + 1 = 0

I pulled an x^2 out:

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

So, a root would be positive one...

Ruby: You somehow converted a five-term polynomial into a four-term polynomial (since x²(x² - x) - (x - 1) = x⁴ - x³ - x - 1), but you did not factor x² out of the polynomial. (To factor an x² out, it would have to have been a factor of each term, and clearly x² is not a factor of x or of 1.)

Since roots come from factors, and since neither (-x + 1) nor (x - 1) is a factor (at least not as you have it, since your expression still involves addition, not just multiplication, of the parenthetical terms), there is no way to conclude that x = 1 is a root. Also, if x = 1 were a root of:

. . . . .f(x) = x⁴ - x³ + x² - x + 1

...then, by definition, f(1) would equal zero. But:

. . . . .f(1) = (1)⁴ - (1)³ + (1)² - (1) + 1 = 1 - 1 + 1 - 1 + 1 = 3 - 2 = 1

...not zero.

Your methods and assumptions raise questions: Have you factored quadratics at all? Have you worked with graphing at all, comparing graphs of x-intercepts to lists of roots and/or factors? Have you heard of the Rational Roots Tests?

Thank you.

Eliz.

 

[stapel]

Would someone please show Ruby and I how to find the imaginary roots?

Jwpaine: There are different methods, one of which (specific to this exercise) is explained by sorobon. When you can't find nice rational roots, the problem generally becomes much more difficult, and sometimes you just have to resort to numerical methods.

In this particular case, however, you can find the solution working backwards from some reasonable assumptions.

. . .We have x⁴ - x³ + x² - x + 1, a suspiciously-neat and
. . .simple polynomial. It's a degree-four polynomial.
. . .Let's assume that it's the product of two quadratics.
. . .Then:

. . . . .x⁴ - x³ + x² - x + 1 = (x² + ax + 1)(x² + bx + 1)

. . .The x²'s and the 1's are fairly obvious. Now we need
. . .to try to find values for our variables "a" and "b".
. . .Multiplying out, we get:

. . . . .x⁴ + (a + b)x³ + (ab + 2)x² + (a + b)x + 1

. . .Equating coefficients, we get:

. . . . .ab + 2 = 1
. . . . .ab = -1
. . . . .b = -1/a

. . . . .a + b = 1
. . . . .a + (-1/a) = 1
. . . . .a² - 1 = a
. . . . .a² - a - 1 = 0

Then solve for the value(s) of "a", back-solve for the corresponding value(s) of "b", and plug-n-chug. Then apply the Quadratic Formula to each of the quadratic factors to find the (complex) results provided earlier.

I hope that helps a bit.

Eliz.