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

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

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Answer 1 · 2018-04-05T04:56:48

$9 y^{2} + 12 y + 4$ $9y^2+12y+4$

Explanation:

When a polynomial is squared, that means you have to multiply it by itself.
${(3 y + 2)}^{2}$ $(3y+2)^2$
$(3 y + 2) (3 y + 2)$ $(3y+2)(3y+2)$

Multiply the terms in the first polynomial by the ones in the second polynomial and combine like terms.
$(3 y + 2) (3 y + 2)$ $(3y+2)(3y+2)$
$9 y^{2} + 6 y + 6 y + 4$ $9y^2+6y+6y+4$
$9 y^{2} + 12 y + 4$ $9y^2+12y+4$

Answer 2 · 2018-04-05T05:24:21

${(3 y + 2)}^{2} = 9 y^{2} + 12 y + 4$ $(3y+2)^2=color(blue)(9y^2+12y+4$

Explanation:

Multiply/Simplify/Expand:

${(3 y + 2)}^{2}$ $(3y+2)^2$

Use the square of a sum:

${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2=a^2+2ab+b^2$ ,

where:

$a = 3 y$ $a=3y$ , and $b = 2$ $b=2$ .

Plug in the known values.

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

Simplify ${(3 y)}^{2}$ $(3y)^2$ to $9 y^{2}$ $9y^2$ .

${(3 y + 2)}^{2} = 9 y^{2} + 2 \cdot 3 y \cdot 2 + 2^{2}$ $(3y+2)^2=9y^2+2*3y*2+2^2$

Simplify $2^{2}$ $2^2$ to $4$ $4$ .

${(3 y + 2)}^{2} = 9 y^{2} + 2 \cdot 3 y \cdot 2 + 4$ $(3y+2)^2=9y^2+2*3y*2+4$

Simplify $2 \cdot 3 y \cdot 2$ $2*3y*2$ to $12 y$ $12y$ .

${(3 y + 2)}^{2} = 9 y^{2} + 12 y + 4$ $(3y+2)^2=9y^2+12y+4$

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

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