Find all root :
$p(x)=x^4+2x^3-x^2-2x+7=0$
Where $p(\alpha)=0$ ,
$\alpha=\sqrt{2}+\omega$
$\omega=e^{\frac{2iπ}{3}}$
My try :
Since : $\alpha$ root of equation then
$\bar\alpha$
also root
But I don't know how I complete to find all root
Answer is :
$x_1=\alpha$
$x_2=\sqrt{2}+\bar\omega$
$x_3=-\sqrt{2}+\bar\omega$
$x_2=-\sqrt{2}+\omega$
On
Usually, the first step in solving an equation of fourth degree is to clear by substitution the term in $x^3$. In our case, this is done by substituting $x+\frac 12=y$. So $x= y-\frac 12$, we plug in this value and get succesively $$ \begin{aligned} f(x) &= x^4 + 2x^3-x^2-2x+7\ ,\\ g(y) &:= f(y-\frac 12) \\ &= y^4 -\frac 52 y^2+\frac {121}{16} \ . \end{aligned} $$ Now we have luck, we get a biquadratic equation, so $$ \begin{aligned} y^2 &= \frac 12\left(\frac 52\pm \sqrt{\left(\frac 52\right)^2-4\cdot\frac {121}{16}}\right) \\ &= \frac 12\left(\frac 52\pm \sqrt{-24}\right) =\frac 14(5\pm 4\sqrt{-6}) =\frac 54 \pm \sqrt 6 \\ &= \left(\frac{\sqrt{-3}}2\pm\sqrt 2\right)^2\ . \end{aligned} $$ So the $y$--roots of the polynomial $g$ are $$ \pm \frac{\sqrt{-3}}2\pm\sqrt 2\ , $$ and the $x$--roots of $f$ are: $$ -\frac 12\pm \frac{\sqrt{-3}}2\pm\sqrt 2\ . $$
Computer check, sage:
sage: var('x,y');
sage: f(x) = x^4 + 2*x^3 - x^2 - 2*x + 7
sage: g(y) = f(y-1/2).expand()
sage: g(y)
y^4 - 5/2*y^2 + 121/16
sage: g.roots(multiplicities=False)
[-1/2*sqrt(4*I*sqrt(6) + 5),
1/2*sqrt(4*I*sqrt(6) + 5),
-1/2*sqrt(-4*I*sqrt(6) + 5),
1/2*sqrt(-4*I*sqrt(6) + 5)]
sage: f.roots(multiplicities=False)
[-1/2*I*sqrt(3) - sqrt(2) - 1/2,
-1/2*I*sqrt(3) + sqrt(2) - 1/2,
1/2*I*sqrt(3) - sqrt(2) - 1/2,
1/2*I*sqrt(3) + sqrt(2) - 1/2]
As Shubham Joni has already mentioned in a comment, if you know two roots $\alpha$ and $\bar\alpha$ of this polynomial, you can decompose it in the form $$ p(x) = (x-\alpha)(x-\bar\alpha) q(x)$$ where $q(x)$ is a polynomial of second degree, and its roots can be found through its discriminant. To find $q(x)$ itself you can use polynomial long division method.