Answer
$$\frac{x^2+4x+4}{3}x^2=\frac{x^4+4x^3+4x^2}{3}$$
Explanation
| $$ \begin{aligned}\frac{x^2+4x+4}{3}x^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^4+4x^3+4x^2}{3}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{x^2+4x+4}{3} $ by $ x^2 $ to get $ \dfrac{ x^4+4x^3+4x^2 }{ 3 } $. Step 1: Write $ x^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{x^2+4x+4}{3} \cdot x^2 & \xlongequal{\text{Step 1}} \frac{x^2+4x+4}{3} \cdot \frac{x^2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( x^2+4x+4 \right) \cdot x^2 }{ 3 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^4+4x^3+4x^2 }{ 3 } \end{aligned} $$ |
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