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

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Answer 1 · 2015-03-24T12:59:12

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

Plug that in:

$= {(3 x)}^{2} + 2 \cdot (3 x) (- 2 y) + {(- 2 y)}^{2}$ $=(3x)^2+2*(3x)(-2y)+(-2y)^2$

$= 9 x^{2} - 12 x y + 4 y^{2}$ $=9x^2-12xy+4y^2$

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

$= (3 x - 2 y) \cdot (3 x - 2 y)$ $=(3x-2y)*(3x-2y)$

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

$= 3 x \cdot 3 x + 3 x \cdot (- 2 y) + (- 2 y) \cdot 3 x + (- 2 y) \cdot (- 2 y)$ $=3x*3x+3x*(-2y)+(-2y) * 3x+(-2y)*(-2y)$

Which (of course) wil give the same result.

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