How do you multiply ${(3 x + 9)}^{2}$ $(3x+9)^2$ ?

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How do you multiply ${(3 x + 9)}^{2}$ $(3x+9)^2$ ?

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Answer 1 · 2015-07-03T12:40:49

I found: $9 x^{2} + 54 x + 81$ $9x^2+54x+81$

Explanation:

Your square can be "translated" as:
${(3 x + 9)}^{2} = (3 x + 9) (3 x + 9) =$ $(3x+9)^2=(3x+9)(3x+9)=$
multiplying we get:
$= (3 x \cdot 3 x) + (3 x \cdot 9) + (9 \cdot 3 x) + (9 \cdot 9) =$ $=(3x*3x)+(3x*9)+(9*3x)+(9*9)=$
$= 9 x^{2} + 27 x + 27 x + 81 =$ $=9x^2+27x+27x+81=$
and finally:
$= 9 x^{2} + 54 x + 81$ $=9x^2+54x+81$

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How do you multiply ${(3 x + 9)}^{2}$ $(3x+9)^2$ ?

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