How to factorize $(x-2)^5+x-1$? This is a difficult problem. How to factorize this? $$(x-2)^5+x-1$$ we can't do any thing now and we should expand it first: $$x^5-10x^4+40x^3-80x^2+81x-33$$ but I can't factorize it.

Using the substitution $t=x-2$, it suffices to factor $t^5 + t+1$.

But $t^2+t+1$ divides $t^5+t+1$. Why?

$t^2+t+1 = (t-\zeta)(t-\zeta')$, where $\zeta,\zeta'$ are the primitive $3rd$ roots of unity, and thus $\zeta^5 + \zeta + 1 = \zeta^2 + \zeta + 1 = 0$, similarly for $\zeta'$.

Now, the quotient $(t^5+t+1)/(t^2+t+1)$ is easily determined via polynomial division.

_Alternative solution._ Using $t^3-1=(t^2+t+1)(t-1)$, we get: $$t^5+t+1=(t^5-t^2)+(t^2+t+1)=(t^3-1)t^2 + (t^2+t+1)$$ $$=(t^2+t+1)(t-1)t^2 + (t^2+t+1) = (t^2+t+1)((t-1)t^2+1)$$