Factor completely over integer coefficients

[Perhaps check it again]

x⁷ + x⁶ + x⁵ + x⁴ + x³ + x² + x + 1

(Please do it without using the assistance of a computer, calculator, or mathematics formula book.)

[superprismatic]

Most mathematically minded people know that if P(x) is a polynomial in x

then r is a root of P if and only if x-r is a factor of P(x). More generally,
replacing every instance of xⁿ in P(x) by r results in creating the trivial
polynomial 0 if and only if xⁿ-r is a factor of P(x).



Since we wish to annihilate that pesky constant term 1 when we replace things in P(x), it's natural to look to roots of -1 thusly:

Let P(x)=x⁷+x⁶+x⁵+x⁴+x³+x²+x+1. If x were equal to -1, then P(x) becomes
-1+1-1+1-1+1-1+1 which is identically 0. This means that x-(-1), i.e. x+1, is a factor.

Now, P(x)=x·(x²)³+(x²)³+x·(x²)²+(x²)²+x·(x²)+x²+x+1. So if x² were to equal -1, then P(x)
would simplify to -x-1+x+1-x-1+x+1 which is 0. So x²-(-1), i.e. x²+1, is another factor.

For x³=-1, we get P(x)=x·(x³)²+(x³)²+x²·x³+x·x³+x³+x²+x+1 which simplifies to
-x+1-x²-x-1+x²+x+1 or 1-x which is not identically 0, so x³+1 is not a factor.

For x⁴=-1, we get P(x)=-x³-x²-x-1+x³+x²+x+1 which is identically 0, so x⁴+1 is another factor.

Since the product of these three factors is of degree 7, we are done.

So, x⁷+x⁶+x⁵+x⁴+x³+x²+x+1 = (x+1)·(x²+1)·(x⁴+1)