How to solve $2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0$?

Question

How to solve $$2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0?$$

I just have no idea and I'have some knowledge about the polynomial equations. Here, just nothing.

Answer 1

Substitute $x = \sqrt{2}y$ and divide by $\sqrt{2}^5$. The resulting equation is

$$y^5 + (y+1)^5 = 0.$$

Now substitute $y = -1/(z+1)$ and multiply by $(z+1)^5$ and you get $$z^5 - 1 = 0.$$

Answer 2

The only real root is $$ \alpha=-\frac{1}{\sqrt{2}}=- 0.707106781187\cdots $$

The polynomial divided by $x-\alpha$ then has degree $4$ and has roots $\alpha\pm i\beta$, $\alpha\pm i \gamma$.

Answer 3

We had one root $x=-\frac{1}{\sqrt{2}}$ as already said in answers and comments. After long division remains $$2 x^4+4 \sqrt{2} x^3+16 x^2+12 \sqrt{2} x+8=0$$ Make $x=y-\frac{1}{\sqrt{2}}$ to get $$2 y^4+10 y^2+\frac{5}{2}=0$$ which is a quadratic in $y^2$.