How i can solve this square of binomial (3z+2k)2?

asked Nov 15, 2015 161k views

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You can solve this by the 1st identity which is (a+b)=a^2+b^2+2ab
(3z)^2+(2k)^2+2*3z*2k
=9z^2+4k^2+12kz

IronBlossom answered Nov 15, 2015

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3 votes

$(3z+2k)^2=(3z)^2+2\cdot 3z\cdot 2k +(2k)^2 = 9z^2 + 12kz +4k^2$

JessieArr answered Nov 20, 2015

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