how i can solve this square of binomial (3z+2k)2?
You can solve this by the 1st identity which is (a+b)=a^2+b^2+2ab(3z)^2+(2k)^2+23z2k=9z^2+4k^2+12kz
\((3z+2k)^2=(3z)^2+2\cdot 3z\cdot 2k +(2k)^2 = 9z^2 + 12kz +4k^2\)
