How do you factor completely #m^8 +m^4/16 + 1/256#?

1 Answer
Sep 17, 2016

Answer:

#m^8+m^4/16+1/256#

#= (m^2-sqrt(3)/2m+1/4)(m^2+sqrt(3)/2m+1/4)(m^2-1/2m+1/4)(m^2+1/2m+1/4)#

Explanation:

Note that:

#(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4#

Let #n = 1/2#

Then:

#m^8+m^4/16+1/256#

#= (m^2)^4+(m^2)^2(n^2)^2+(n^2)^4#

#= ((m^2)^2-(m^2)(n^2)+(n^2)^2)((m^2)^2+(m^2)(n^2)+(n^2)^2)#

#= (m^4-m^2n^2+n^4)(m^4+m^2n^2+n^4)#

#= (m^2-sqrt(3)mn+n^2)(m^2+sqrt(3)mn+n^2)(m^2-mn+n^2)(m^2+mn+n^2)#

#= (m^2-sqrt(3)/2m+1/4)(m^2+sqrt(3)/2m+1/4)(m^2-1/2m+1/4)(m^2+1/2m+1/4)#