Factor x5 + x4 + x3 + x2 + x + 1 completely.

Step-by-step solution

1. 1

   Recognise a geometric sum. The six terms are $x^0$ through $x^5$ with ratio $x$, so for $x\neq 1$

   $1+x+x^2+x^3+x^4+x^5=\frac{x^6-1}{x-1}$

   This converts an awkward quintic into a difference of powers, which has known factorizations.

2. 2

   Factor $x^6-1$ as a difference of cubes of squares. Reading $x^6=(x^3)^2$ gives

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

3. 3

   Factor each cubic with the sum and difference of cubes.

   $x^3-1=(x-1)(x^2+x+1),\qquad x^3+1=(x+1)(x^2-x+1)$

4. 4

   Cancel the $x-1$. Substituting everything back,

   $\frac{(x-1)(x^2+x+1)(x+1)(x^2-x+1)}{x-1}=(x+1)(x^2+x+1)(x^2-x+1)$

   The cancellation is legitimate as an identity of polynomials — the original quintic has no factor $x-1$, since substituting $x=1$ gives $6$, not $0$.

5. 5

   Cross-check by direct grouping. Pairing terms, $x^4(x+1)+x^2(x+1)+(x+1)=(x+1)(x^4+x^2+1)$, and $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$ (add and subtract $x^2$ to make a difference of squares). Same answer by a completely different route.

6. 6

   Confirm numerically and check irreducibility. At $x=2$: the original is $32+16+8+4+2+1=63$, and $(3)(7)(3)=63$ ✓. Each quadratic has discriminant $1-4=-3<0$, so neither factors over the reals — the factorization is complete.