3[tex] \sqrt{10} [/tex] ( y^2*[tex] \sqrt{4} [/tex]+[tex] \sqrt{8y} [/tex] )
3y^2[tex] \sqrt{40} [/tex]+3[tex] \sqrt{80y} [/tex]
[tex] \sqrt{40} [/tex] = [tex] \sqrt{2*2*10} [/tex] = 2[tex] \sqrt{10} [/tex]
[tex] \sqrt{80y} [/tex] = 4[tex] \sqrt{5y} [/tex]
6[tex] y^{2} [/tex][tex] \sqrt{10} [/tex]+12[tex] \sqrt{5y} [/tex]
The answer is the first picture that contains an answer
