How do you multiply #(3x-2y)^2#?

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Answer 1 · 2015-03-24T12:59:12

The special product #(A+B)^2=A^2+2AB+B^2#
where (in this case) #A=3x# and #B=-2y#

Plug that in:

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

#=9x^2-12xy+4y^2#

Or:
If you forgot the special products, you can expand:

#=(3x-2y)*(3x-2y)#

And then work using FOIL, which means:
First - Outer - Inner -Last

#=3x*3x+3x*(-2y)+(-2y) * 3x+(-2y)*(-2y)#

Which (of course) wil give the same result.