Roots over $\mathbb{C}$ equation $x^{4} - 4x^{3} + 2x^{2} + 4x + 4=0 $.

Question

I need roots over $\mathbb{C}$ equation $$x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = 0$$

From Fundamental theorem of algebra we have statement that the equation have 4 roots over complex.

But I prepare special reduction: $$ \color{red}{ x^{4} - 4x^{3} + 2x^{2} + 4x + 4} = (x-1)^{4}-4(x-1)^{2} + 7 $$ for substitution $y = (x-1)^{2}$ we have: $y^{2} - 4y + 7 = 0 $

$y_{0} = 2+i\sqrt{3}$

$y_1 = 2 - i\sqrt{3}$

and we have
$y_{0}^{1/2} + 1 = x_{0} $, $-y_{0}^{1/2} + 1 = x_{1}$,
$y_{1}^{1/2} + 1 = x_{2}$, $-y_{1}^{1/2} +1 = x_{3} $

I am not sure what is good results. Please check my solution.

EDIT: The LHS is not correct, I modify this equation. We should have $p(x) = x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = (x-1)^{4}-4(x-1)^{2} + 7 $

EDIT2: I need show that the $p(x)$ is reducible (or not) over $\mathbb{R}[x]$ for two polynomials of degrees 2. But I am not sure how show that $\left(x-1-\sqrt{2-i\sqrt{3} }\right) \left(x-1+\sqrt{(2+i\sqrt3}\right)$ is (not) polynomial of degree 2.

Answer 1

COMMENT.-These are the two real roots given by Wolfram. It is impossible that you can calculate them by simple means. The two non-real roots are equally complicated.

Answer 2

The equality that you have used is not correct.$$x^{4}-4x^{3}+2x^{2}+4 \ne (x-1)^{4}-4(x-1)^{2} + 7$$

Answer 3

Solving algebraically is perhaps not too messy if one is happy to have roots in terms of the solution to a cubic:-

Using the substitution $t=x-1$ we have $$ t^{4}- 4t^{2}-4t+3=0$$ and therefore $$ (t^{2}- 2)^2=4t+1.$$

For any $z$, $$ (t^{2}- 2+2z^2)^2=4z^2t^2+4t+4z^4-8z^2+1.$$

Let $z$ be a solution (found by Cardan's method) of the cubic in $z^2$ $$4z^6-8z^4+z^2-1=0$$ Then $$ (t^{2}- 2+2z^2)^2=(2zt+\frac{1}{z})^2.$$ This is a quadratic in $t$ giving, for example, $t=z+\sqrt {2-z^2+\frac{1}{z}}$ and $t=z-\sqrt {2-z^2+\frac{1}{z}}$.

(One has to be careful choosing the correct signs for whatever root of the cubic is chosen- the signs I have used give real roots for the real root of the cubic.)

Answer 4

Answer to the revised question

You have correctly obtained $x=1\pm \sqrt{2\pm i\sqrt3}$.

By the standard method find the complex square roots. This gives one form of the full answer as $$x=1\pm \sqrt[4]7(\frac{2+\sqrt7\pm i\sqrt3}{\sqrt{14+4\sqrt7}}).$$