How do you factor #x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x +1#?

1 Answer
May 11, 2016

Answer:

#x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1#

#= (x^2+x+1)(x^2-2cos((2 pi)/9)x+1)(x^2-2cos((4 pi)/9)x+1)(x^2-2cos((8 pi)/9)x+1)#

Explanation:

Notice that if you multiply this polynomial by #(x-1)# the result is #x^9-1#.

So the zeros of this polynomial are all the Complex #9#th roots of #1# except #1#.

Using the difference of cubes identity a couple of times, we find:

#x^9-1 = (x^3)^3-1^3 = (x^3-1)(x^6+x^3+1) = (x-1)(x^2+x+1)(x^6+x^3+1)#

Then we can divide both ends by #(x-1)# to get:

#x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 = (x^2+x+1)(x^6+x^3+1)#

No surprise there.

Note that the zeros of #x^2+x+1# are the Complex #3#rd roots of #1#.

The zeros of #x^6+x^3+1# are the Complex #9#th roots that are not #3#rd roots, i.e.

#e^((2k pi)/9i) = cos((2k pi)/9) + i sin((2k pi)/9)#

with #k = +-1, +-2, +-4#

The factors with Real coefficients, composed of Complex conjugate pairs are:

#(x-cos((2k pi)/9)-i sin((2k pi)/9))(x-cos((2k pi)/9)+i sin((2k pi)/9))#

#=x^2-2cos((2k pi)/9)x+1#

for #k = 1, 2, 4#

Hence:

#x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1#

#= (x^2+x+1)(x^2-2cos((2 pi)/9)x+1)(x^2-2cos((4 pi)/9)x+1)(x^2-2cos((8 pi)/9)x+1)#