Showing a polynomial is solvable by radicals

[PsychonautQQ]

Homework Statement

Show that the polynomial f(x) = x^5 - x^3 - 3x^2 + 3 is solvable by radicals where the coefficients of f are from the field of rational numbers.

Homework Equations

The Attempt at a Solution

My strategy to solve this problem was to construct a splitting field and then see if that splitting field lies in some radical extension. I first noted that 1 is a root of f(x), so I divided f(x) by (x-1) and got the quotient x^4 + x^3 - 3x - 3.

From here I hit a block. I tried assuming r to be a root and then divided x^4 + x^3 - 3x - 3 by (x-r) to arrive at the quotient x^3 + (r+1)x^2 + (r^2+r)x + (r^3 + r^2 - 3) which didn't seem to be much help. How else can I try to find the roots of x^4 + x^3 - 3x - 3 so I can construct a splitting field?

 

[lurflurf]

factor by grouping
$$x^5 - x^3 - 3x^2 + 3=x^3(x^2-1)-3(x^2-1)$$

 

[Ray Vickson]

Homework Statement

Show that the polynomial f(x) = x^5 - x^3 - 3x^2 + 3 is solvable by radicals where the coefficients of f are from the field of rational numbers.

Homework Equations

The Attempt at a Solution

My strategy to solve this problem was to construct a splitting field and then see if that splitting field lies in some radical extension. I first noted that 1 is a root of f(x), so I divided f(x) by (x-1) and got the quotient x^4 + x^3 - 3x - 3.

From here I hit a block. I tried assuming r to be a root and then divided x^4 + x^3 - 3x - 3 by (x-r) to arrive at the quotient x^3 + (r+1)x^2 + (r^2+r)x + (r^3 + r^2 - 3) which didn't seem to be much help. How else can I try to find the roots of x^4 + x^3 - 3x - 3 so I can construct a splitting field?

One root of $x^4 + x^3 - 3x - 3$ is $x = -1$. Alternatively, you can use classical formulas for the roots of a quartic polynomial.

 

[PsychonautQQ]

factor by grouping
$$x^5 - x^3 - 3x^2 + 3=x^3(x^2-1)-3(x^2-1)$$

Nice find. Could you elaborate on how this grouping will help me find all 5 roots? Doing this definitely helps me to see that (+/-)1 are both roots, but now what?

 

[PsychonautQQ]

One root of $x^4 + x^3 - 3x - 3$ is $x = -1$. Alternatively, you can use classical formulas for the roots of a quartic polynomial.

Ah, how could I have missed that. Thank you, now this polynomial factors as (x-1)(x+1)(x^3 - 3) where x^3-3 has complex roots, and therefore I believe this polynomial will not be solvable by radicals

 

[fresh_42]

Ah, how could I have missed that. Thank you, now this polynomial factors as (x-1)(x+1)(x^3 - 3) where x^3-3 has complex roots, and therefore I believe this polynomial will not be solvable by radicals

What makes you believe this? Can't we solve equations of degree $3$? And what is the algebraic reason?

 

[Ray Vickson]

Ah, how could I have missed that. Thank you, now this polynomial factors as (x-1)(x+1)(x^3 - 3) where x^3-3 has complex roots, and therefore I believe this polynomial will not be solvable by radicals

If you think (wrongly) that a radical must be of the form $\sqrt{\cdots}$, then your conclusion is valid. However, radicals are things like $\sqrt[n]{\cdots} = (\cdots)^{1/n}$ for integer $n$, so certainly the equation $x^3 = 3$ is solvable by radicals.

 

[epenguin]

factor by grouping
$$x^5 - x^3 - 3x^2 + 3=x^3(x^2-1)-3(x^2-1)$$

I thought of that too - however it looks to me from previous related questions that the questions are not so much to do with such base aims as solving equations as formulating the maths in some more hifalutin' abstract algebra language.

Here it looks like setting that fairly elementary procedure that you and I can manage in some wider context. I do not wish to be anti-intellectual. Although I do deal with some applications of polynomials I somehow doubt this will ever be necessary for me, but I can suppose that it has some advantages. It would be interesting to be told what these are. I expect somebody knows - though I wonder if the student knows?

Well from posts it looks like the student knows things we don't know, - but is fairly new and shaky on things that a lot of us know.