Question

Factor the expression completely over the complex numbers.

y4−16

Answer 1

i'm not sure about this answer but hope it works (y^2+4)(y+2)(y-2)

Answer 2

`\bf \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b) \\\\\\ \textit{and recall that }i^2=-1\\\\ -------------------------------\\\\ y^4-16\implies (y^2)^2-4^2\implies (y^2-4)(y^2+4)\\\\\\ (y^2-2^2)(y^2+4) \implies (y-2)(y+2)(y^2+4)\quad \begin{cases} +4=-(-2^2)\\ \qquad -(-1\cdot 2^2)\\ \qquad -(i^2\cdot 2^2)\\ \qquad -(2^2i^2)\\ \qquad -(2i)^2 \end{cases} \\\\\\ (y-2)(y+2)[y^2-(2i)^2]\implies (y-2)(y+2)[(y-2i)(y+2i)] \\\\\\ (y-2)(y+2)(y-2i)(y+2i)`