Perhaps check it again 22 Report post Posted January 26, 2014 x^{7} + x^{6 } + x^{5 }+ x^{4} + x^{3 }+ x^{2} + x + 1 (Please do it without using the assistance of a computer, calculator, or mathematics formula book.) Share this post Link to post Share on other sites
0 superprismatic 11 Report post Posted January 27, 2014 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 x^{n} in P(x) by r results in creating the trivial polynomial 0 if and only if x^{n}-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)=x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+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·(x^{2})^{3}+(x^{2})^{3}+x·(x^{2})^{2}+(x^{2})^{2}+x·(x^{2})+x^{2}+x+1. So if x^{2} were to equal -1, then P(x) would simplify to -x-1+x+1-x-1+x+1 which is 0. So x^{2}-(-1), i.e. x^{2}+1, is another factor. For x^{3}=-1, we get P(x)=x·(x^{3})^{2}+(x^{3})^{2}+x^{2}·x^{3}+x·x^{3}+x^{3}+x^{2}+x+1 which simplifies to -x+1-x^{2}-x-1+x^{2}+x+1 or 1-x which is not identically 0, so x^{3}+1 is not a factor. For x^{4}=-1, we get P(x)=-x^{3}-x^{2}-x-1+x^{3}+x^{2}+x+1 which is identically 0, so x^{4}+1 is another factor. Since the product of these three factors is of degree 7, we are done. So, x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1 = (x+1)·(x^{2}+1)·(x^{4}+1) Share this post Link to post Share on other sites
x^{7} + x^{6 } + x^{5 }+ x^{4} + x^{3 }+ x^{2} + x + 1
(Please do it without using the assistance of a computer, calculator, or mathematics formula book.)
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