How to factorize $(x-2)^5+x-1$?

Question

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.

Answer 1

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: $$t^5+t+1=(t^5-t^2)+(t^2+t+1)=(t^3-1)t^2 + (t^2+t+1)$$ $$=(t^2+t+1)(t-1)t^2 + (t^2+t+1) = (t^2+t+1)((t-1)t^2+1)$$

Answer 2

you can also use $$(x-2)^5 + x - 1 = (x-2)^5 + (x-2) + 1=(x-2+0.754)(x^4+\cdots) $$

$\bf edit:$ $0.754$ is the only real root of $t^5+t+1 = 0.$

Answer 3

finally found how looking in my previous questions.

first substitute $n=x-2$ then

$$(x-2)^5+x-1=n^5+n+1$$

$$n^5+n+1=n^5-n^2+n^2+n+1=n^2(n^3-1)+n^2+n+1$$

$$=n^2(n-1)(n^2+n+1)+n^2+n+1$$

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