How do you factor #x^12+8#?

1 Answer
Feb 16, 2016

Answer:

Use some identities to find the quadratic factors of #x^12+8#

Explanation:

First notice that #x^12 >= 0# for any Real value of #x#. So #x^12+8 >= 8 > 0# for any Real #x# and this polynomial has no linear factors with Real coefficients.

We can find all of the quadratic factors as follows:

The sum of cubes identity can be written:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

Hence:

#x^12+8=(x^4)^3+2^3#

#=(x^4+2)((x^4)^2-(x^4)(2)+2^2)#

#=(x^4+2)(x^8-2x^4+4)#

Notice that:

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

In particular, if #k=sqrt(2)# then we find:

#(a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2)=a^4+b^4#

Hence (putting #a=x# and #b=root(4)(2)#) we find:

#(x^4+2) = (x^2-root(4)(8)x+sqrt(2))(x^2+root(4)(8)x+sqrt(2))#

If instead we put #k=sqrt(3)# then we find:

#(a^2-sqrt(3)ab+b^2)(a^2+sqrt(3)ab+b^2) = a^4-a^2b^2+b^4#

So if we let #a=x^2# and #b=sqrt(2)# then we find:

#(x^4-sqrt(6)x^2+2)(x^4+sqrt(6)x^2+2) = x^8-2x^4+4#

Putting #k=sqrt(2+sqrt(3))#, #a=x# and #b=root(4)(2)# we find:

#(x^2-sqrt(2+sqrt(3))root(4)(2)x+sqrt(2))(x^2+sqrt(2+sqrt(3))root(4)(2)x+sqrt(2))#

#=x^4-sqrt(6)x^2+2#

Putting #k=sqrt(2-sqrt(3))#, #a=x# and #b=root(4)(2)# we find:

#(x^2-sqrt(2-sqrt(3))root(4)(2)x+sqrt(2))(x^2+sqrt(2-sqrt(3))root(4)(2)x+sqrt(2))#

#=x^4+sqrt(6)x^2+2#

Putting this all together, we find:

#x^12+8 = (x^2-root(4)(8)x+sqrt(2))(x^2+root(4)(8)x+sqrt(2))(x^2-sqrt(2+sqrt(3))root(4)(2)x+sqrt(2))(x^2+sqrt(2+sqrt(3))root(4)(2)x+sqrt(2))(x^2-sqrt(2-sqrt(3))root(4)(2)x+sqrt(2))(x^2+sqrt(2-sqrt(3))root(4)(2)x+sqrt(2))#