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

##### 1 Answer

#### Answer:

#= (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

So the zeros of this polynomial are all the Complex

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^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

The zeros of

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

with

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

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)#