How can you factor $x^4-6x^3+8x^2+2x-1?$

The original question is:

Solve this equation for x:

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

I expanded and simplified it to get $$x^4-6x^3+8x^2+2x-1=0$$

Since neither -1 nor 1 are factors, it appears that the roots will be fractional. How can I proceed from here? Thanks.

Solution 1:

Note that initial equation have following form: $$f(f(x))=x$$ where $f(x) = x^2 -3x+1$

So roots of $f(x)=x$ equation will be roots of your initial equation. Solve this quadratic equation and you will get 2 roots. Other 2 roots you can get from 4-degree polynom factorization.

Solution 2:

Hint: $f(x)=x^2(x-3)^2-(x-1)^2$