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Factor completely over integer coefficients

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x7 + x6 + x5 + x4 + x3 + x2 + x + 1

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

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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)
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