How do you find the product #(3c+4d)^2#?

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Answer 1 · 2017-01-24T06:10:48

#(3c+4d)^2=9c^2+24cd+16d^2#

#(3c+4d)^2#

= #(3c+4d)xx(3c+4d)#

= #3cxx(3c+4d)+4d xx(3c+4d)#

= #3cxx3c+3cxx4d+4d xx 3c+4d xx 4d#

= #9c^2+12cd+12cd+16d^2#

= #9c^2+24cd+16d^2#

Answer 2 · 2017-01-24T06:14:10

#(3c+4d)^2 = 9c^2+24cd+16d^2#

Using the special product #(a+b)^2 = a^2+2ab+b^2#, we can set #a=3c# and #b=4d# to get

#(3c+4d)^2 = (3c)^2 + 2(3c)(4d) + (4d)^2#

#=3^2c^2 + (2)(3)(4)cd + 4^2d^2#

#=9c^2+24cd+16d^2#