How do you multiply #(7x+3) (7x+3)#?

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Answer 1 · 2015-08-21T03:32:06

#(7x+3)(7x+3)=(7x+3)^2=49x^2+42x+9#

#(7x+3)(7x+3)=(7x+3)^2#

#(7x+3)^2# is the square of a sum with the form #(a+b)^2=a^2+2ab+b^2#, where #a=7x# and #b=3#.

#(7x+3)^2=(7x)^2+(2*7x*3)+(3)^2##=#

#(7x+3)^2=49x^2+42x+9#

You could also use the FOIL method.

#(7x+3)(7x+3)=(7x*7x)+(7x*3)+(3*7x)+(3*3)##=#

#(7x+3)(7x+3)=49x^2+21x+21x+9##=#

#(7x+3)(7x+3)=49x^2+42x+9#