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

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

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Answer 1 · 2017-02-23T04:49:24

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

Explanation:

${(3 x - 2)}^{2}$ $(3x-2)^2$ can be expanded to $(3 x - 2) (3 x - 2)$ $(3x-2)(3x-2)$
From here you can use the FOIL method to expand.

The "first" would be $3 x \cdot 3 x = 9 x^{2}$ $3x*3x=9x^2$
The "outside" would be $3 x \cdot - 2 = - 6 x$ $3x*-2=-6x$
The "inside," $- 2 \cdot 3 x = - 6 x$ $-2*3x=-6x$
And the "last," $- 2 \cdot - 2 = 4$ $-2*-2=4$

Putting the terms together you get
$9 x^{2} - 6 x - 6 x + 4$ $9x^2-6x-6x+4$

And combining like terms
$9 x^{2} - 12 x + 4$ $9x^2-12x+4$

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

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