Answer
$$\frac{8x^6-4x^2}{4}x^2=\frac{8x^8-4x^4}{4}$$
Explanation
| $$ \begin{aligned}\frac{8x^6-4x^2}{4}x^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{8x^8-4x^4}{4}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{8x^6-4x^2}{4} $ by $ x^2 $ to get $ \dfrac{ 8x^8-4x^4 }{ 4 } $. 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. $$ \begin{aligned} \frac{8x^6-4x^2}{4} \cdot x^2 & \xlongequal{\text{Step 1}} \frac{8x^6-4x^2}{4} \cdot \frac{x^2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 8x^6-4x^2 \right) \cdot x^2 }{ 4 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 8x^8-4x^4 }{ 4 } \end{aligned} $$ |
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