How to solve $x^4-2x^3-x^2+2x+1=0$?

Solution 1:

Another way to notice the factorization $$x^4-2x^3-x^2+2x+1=0$$ Since $x=0$ is not the root of the equation, divide by $x^2$ to get $$x^2 -2x-1 + \frac{2}{x} + \frac{1}{x^2} = 0$$ Rewrite it as $$x^2+\frac{1}{x^2} - 2\left(x-\frac{1}{x}\right) - 1 = 0$$ or $$\left(x-\frac1x\right)^2 + 2 - 2\left(x-\frac{1}{x}\right) - 1 = 0$$ Substitute $t = x - 1/x$ $$t^2 + 2 - 2t - 1 = 0\\ t^2 - 2t + 1 = 0 \\ (t-1)^2 = 0$$ Substitute back to get the final result $$\left(x - \frac{1}{x} - 1\right)^2 = 0$$ which says $$(x^2-x-1)^2 = 0$$

Solution 2:

Hints:

Let $x-2=p.$

Then you will have a perfect square. It comes from a general fact:

"One more than a product of four consecutive positive integers is a perfect square."

Please take a look at the following link:

Prove that the product of four consecutive positive integers plus one is a perfect square