How do you multiply ${(4 c + 9 d)}^{2}$ $(4c+9d)^2$ ?

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How do you multiply ${(4 c + 9 d)}^{2}$ $(4c+9d)^2$ ?

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Answer 1 · 2017-08-13T21:18:23

See a solution process below:

Explanation:

We can use this rule for multiplication to multiply the expression:

${(x + y)}^{2} = x^{2} + 2 x y + y^{2}$ $(color(red)(x) + color(blue)(y))^2 = color(red)(x)^2 + 2color(red)(x)color(blue)(y) + color(blue)(y)^2$

Substituting $4 c$ $color(red)(4c)$ for $x$ $color(red)(x)$ and $9 d$ $color(blue)(9d)$ for $y$ $color(blue)(y)$ gives:

${(4 c + 9 d)}^{2} \Rightarrow {(4 c)}^{2} + (2 \cdot 4 c \cdot 9 d) + {(9 d)}^{2} \Rightarrow$ $(color(red)(4c) + color(blue)(9d))^2 => (color(red)(4c))^2 + (2 * color(red)(4c) * color(blue)(9d)) + (color(blue)(9d))^2 =>$

$16 c^{2} + 72 c d + 81 d^{2}$ $16c^2 + 72cd + 81d^2$

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How do you multiply ${(4 c + 9 d)}^{2}$ $(4c+9d)^2$ ?

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How do you multiply (4c+9d)2(4c+9d)^2?

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How do you multiply ${(4 c + 9 d)}^{2}$ $(4c+9d)^2$ ?