[tex]3\sqrt{10}(y^2\sqrt4+\sqrt{8y})\\\\\text{Use distributive law: a(b+c)=ab+ac, and we know}\ \sqrt4=2\\\\=3\sqrt{10}\cdot2y^2+3\sqrt{10}\cdot\sqrt{8y}\\\\\text{Use}\ \sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\\\\=6y^2\sqrt{10}+3\sqrt{10\cdot8y}=6y^2\sqrt{10}+3\sqrt{80y}=6y^2\sqrt{10}+3\sqrt{16\cdot5y}\\\\=6y^2\sqrt{10}+3\sqrt{16}\cdot\sqrt{5y}\\\\\text{we know}\ \sqrt{16}=4\\\\=6y^2\sqrt{10}+3\cdot4\sqrt{5y}=\boxed{6y^2\sqrt{10}+12\sqrt{5y}\to a.}[/tex]
