Multiply by using the special case products. $$\left( \frac { 2 } { 3 } w + 1 \right) ^ { 2 }$$
A) $\frac { 4 } { 9 } w ^ { 2 } + \frac { 2 } { 3 } w + 1$
B) $\frac { 4 } { 9 } w ^ { 2 } + \frac { 4 } { 3 } w + 1$
C) $\frac { 4 } { 9 } w ^ { 2 } + 1$
D) $\frac { 4 } { 9 } w ^ { 2 } - \frac { 2 } { 3 } w - 1$

Question

Quizplus · Accepted Answer

This is a case of the square of a binomial, $(a+b)^2 = a^2 + 2ab + b^2$. Here, $a = \frac{2}{3}w$ and $b = 1$, so $a^2 = \left(\frac{2}{3}w\right)^2 = \frac{4}{9}w^2$, $2ab = 2\left(\frac{2}{3}w\right)(1) = \frac{4}{3}w$, and $b^2 = 1^2 = 1$. Adding these together gives $\frac{4}{9}w^2 + \frac{4}{3}w + 1$.

Multiply by Using the Special Case Products (23w+1)2\left( \frac { 2 } { 3 } w + 1 \right) ^ { 2 }(32​w+1)2

Multiply by Using the Special Case Products $\left( \frac { 2 } { 3 } w + 1 \right) ^ { 2 }$