Jhamidn
Answered

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

Sagot :

Riya25idn
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
Lilithidn
[tex](3z+2k)^2=(3z)^2+2\cdot 3z\cdot 2k +(2k)^2 = 9z^2 + 12kz +4k^2[/tex]


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