Bi-quadratic equation

Question

Solve for $x$, it has four different solutions:

$$x^4 -2x^3-6x^2-2x+1=0$$

Answer 1

Hint:

Grouping $-2x^3=2x^2(-x),-2x=2(-x)1$

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

Answer 2

Using the rational roots theorem, you find that $-1$ is a root. Furthermore, you obtain by long division that $\;x^4 -2x^3-6x^2-2x+1=(x+1)(x^3-3x2-3x+1)$ and can check that $-1$ is actually a double root: $$x^4 -2x^3-6x^2-2x+1=(x+1)^2(x^2-4x+1).$$

Answer 3

1]1just ssolve the last part, hope this helps