How do you multiply #(3y + 2) ^ { 2}#?

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Answer 1 · 2018-04-05T04:56:48

#9y^2+12y+4#

When a polynomial is squared, that means you have to multiply it by itself.
#(3y+2)^2#
#(3y+2)(3y+2)#

Multiply the terms in the first polynomial by the ones in the second polynomial and combine like terms.
#(3y+2)(3y+2)#
#9y^2+6y+6y+4#
#9y^2+12y+4#

Answer 2 · 2018-04-05T05:24:21

#(3y+2)^2=color(blue)(9y^2+12y+4#

Multiply/Simplify/Expand:

#(3y+2)^2#

Use the square of a sum:

#(a+b)^2=a^2+2ab+b^2#,

where:

#a=3y#, and #b=2#.

Plug in the known values.

#(3y+2)^2=(3y)^2+2*3y*2+2^2#

Simplify #(3y)^2# to #9y^2#.

#(3y+2)^2=9y^2+2*3y*2+2^2#

Simplify #2^2# to #4#.

#(3y+2)^2=9y^2+2*3y*2+4#

Simplify #2*3y*2# to #12y#.

#(3y+2)^2=9y^2+12y+4#