Just do the long division to get
(2y^4 + z^2)(4y^8 - 2y^4z^2 + z^4)
the binomial 8y^12+z^6 can be factored into a binomial and a trinomial. the binomial factor is (2y^4+z^2). whats the trinomial?
2 answers
or, you could recognize it as the sum of two cubes:
8y^12+z^6
= (2y^4)^3 + (z^2)^3
= (2y^4 + z^2)((2y^4)^2 - (2y^4)(z^2) + (z^2)^2)
= (2y^4 + z^2)(4y^8 - 2y^4z^2 + z^4)
8y^12+z^6
= (2y^4)^3 + (z^2)^3
= (2y^4 + z^2)((2y^4)^2 - (2y^4)(z^2) + (z^2)^2)
= (2y^4 + z^2)(4y^8 - 2y^4z^2 + z^4)