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

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

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Answer 1 · 2017-06-06T21:29:24

The answer is $4 x^{2} - 12 x y + 9 y^{2}$ $4x^2-12xy+9y^2$ .

Explanation:

First, we begin by setting it in the form of

$(2 x - 3 y) (2 x - 3 y)$ $(2x-3y)(2x-3y)$

Using FOIL, we multiply the first term with the outer and inner term of $(2 x) (2 x - 3 y)$ $(2x)(2x-3y)$ . If done right, you should arrive to the answer of

$4 x^{2} - 6 x y$ $4x^2-6xy$

Now we do the same for the last term of the first part so $(- 3 y) (2 x - 3 y)$ $(-3y)(2x-3y)$ . The answer should be

$- 6 x y + 9 y^{2}$ $-6xy+9y^2$

So we now have the following terms of

$4 x^{2} - 6 x y - 6 x y + 9 y^{2}$ $4x^2-6xy-6xy+9y^2$

What we do now is simplify (combine like terms). The final answer is

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

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

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