How do you factor #x^12+8#?
1 Answer
Use some identities to find the quadratic factors of
Explanation:
First notice that
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
#(a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2)=a^4+b^4#
Hence (putting
#(x^4+2) = (x^2-root(4)(8)x+sqrt(2))(x^2+root(4)(8)x+sqrt(2))#
If instead we put
#(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
#(x^4-sqrt(6)x^2+2)(x^4+sqrt(6)x^2+2) = x^8-2x^4+4#
Putting
#(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
#(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))#