How to evaluate $x^6+x^4+x^3+x^2+1=0$?

Question

There is a hint in the question, use the factorization of $x^5+x+1$.

Answer 1

Here is a better hint:

$$x^6+x^4+x^3+x^2+1=(x^4+x^3+x^2+x+1)(x^2-x+1)$$

Answer 2

How about first multiplying by $(x-1)$ ?

$$(x-1)(x^6+x^4+x^3+x^2+1)= x^7-x^6+x^5 - x^2+x-1= (x^2-x+1)(x^5-1)$$ and carry on from there?

Added: Not sure if worth mentioning, but doing the same on $x^5+x+1$ requires an extra little trick: $$ (x-1)(x^5+x+1)= x^6-x^5 (+x^3 - x^3) +x^2-1=(x^3-x^2+1)(x^3-1)$$

Answer 3

Divide by $x^3$ and remember that $$ x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right) $$ so the equation becomes $$ \left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right) +\left(x+\frac{1}{x}\right)+1=0 $$ Set $t=x+1/x$ and you get $$ t^3-2t+1=0 $$ Can you go on? Note that $t=1$ is a solution.

Answer 4

I missed something. Standard contest trick. The exponents used are $$ 1,4,3,2,0 \pmod 5 $$ while all coefficients are $1.$

That means that the nontrivial fifth roots of unity are roots and the polynomial must be divisible by $$ x^4 + x^3 + x^2 + x + 1 $$