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

1 Answer
Aug 21, 2015

Answer:

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

Explanation:

#(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.

http://www.mesacc.edu/~scotz47781/mat120/notes/polynomials/foil_method/foil_method.html

#(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#