factor the expression completely over the complex numbers. $y^{4}+12y^{2}+36$ enter your answer in the box.

Explanation:

Step1: Recognize perfect - square trinomial

Let (x = y^{2}), then the expression (y^{4}+12y^{2}+36) becomes (x^{2}+12x + 36). Since (a = 1), (b=12), (c = 36) and (b^{2}-4ac=12^{2}-4\times1\times36=144 - 144=0), and (x^{2}+12x + 36=(x + 6)^{2}) (using the formula ((a + b)^2=a^{2}+2ab + b^{2}) where (a=x) and (b = 6)). Substituting back (x=y^{2}), we get ((y^{2}+6)^{2}).

Step2: Factor over complex numbers

We know that (y^{2}+6=y^{2}-(- 6)). Using the difference - of - squares formula (a^{2}-b^{2}=(a + bi)(a - bi)) (where (a=y) and (b=\sqrt{6})), (y^{2}+6=(y+\sqrt{6}i)(y-\sqrt{6}i)). So ((y^{2}+6)^{2}=(y+\sqrt{6}i)^{2}(y-\sqrt{6}i)^{2}).

Answer:

((y+\sqrt{6}i)^{2}(y - \sqrt{6}i)^{2})