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.

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