Can anyone please help me with this?

1 answer:

6 0

$y^4+12y^2+36=(y^2)^2+12y^2+36\\\\\text{substitute}\ y^2=t\\\\t^2+12t+36=t^2+2\cdot t\cdot6+6^2\leftarrow\text{It's a perfect square}\\\\\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\=(t+6)^2=(t+6)(t+6)\\\\\text{we're going back to substitution}\\\\=(y^2+6)(y^2+6)\\\\(\sqrt6)^2=6\to(i\sqrt6)^2=-6\\\\\text{therefore}\\\\=(y^2-(i\sqrt6)^2)(y^2-(i\sqrt6)^2)\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\boxed{(y-i\sqrt6)(y+i\sqrt6)(y-i\sqrt6)(y+i\sqrt6)}$

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