Multiply: (3x + 3y)2 ?

1)
(3x + 3y)^2
= (3x + 3y)(3x + 3y)
= 3x*3x + 3y*3x + 3x*3y + 3y*3y
= 9x^2 + 9xy + 9xy + 9y^2
= 9x^2 + 18xy + 9y^2

2)
(3x + 4)^2
= (3x + 4)(3x + 4)
= 3x*3x + 4*3x + 3x*4 + 4*4
= 9x^2 + 12x + 12x + 16
= 9x^2 + 24x + 16

(3x+3y)^2=(3x+3y)(3x+3y)
=3x*3x + 3x*3y + 3y*3x + 3y*3y
=9x^2 +9xy + 9yx +9y^2 (note xy=yx)
=9x^2 +18xy +9y^2
=9(x^2 +2xy +y^2)

6x+6y
===================================================
9x^2 + 24x +16

Use the FOIL method.
Multiply the first, outer, inner, then last. :]
So, to work it out:

(3x + 3y)(3x + 3y)
F: 3x*3x= 9x
O:3x*3y=9xy
I:3x*3y=9xy
L:3y*3y=9y

Solution: 3x+18xy+3y

Monica C...just the answer IS not the solution....grasp the concept....
If the problem was (3+3)2, then it convert to (6)2=12
that is similar to multiplying each of the 3 by 2, then sum up...example
(3+3)2=(6+6)=12
Since you now have x and y in your problem, follow the same logic...
multiply each term then sum up.
(3x+3y)2=(6x+6y) same as 6(x+y)
...concept......good luck

6x+6y

6x + 6y.

9x^2 + 24x + 16 for the second one

6x + 6y

brackets should be multiplied by the values outside.