Answer
$$8\cdot\frac{x^3}{x^2+4x}=\frac{8x^3}{x^2+4x}$$
Explanation
| $$ \begin{aligned}8 \cdot \frac{x^3}{x^2+4x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{8x^3}{x^2+4x}\end{aligned} $$ | |
| ① | Multiply $8$ by $ \dfrac{x^3}{x^2+4x} $ to get $ \dfrac{ 8x^3 }{ x^2+4x } $. Step 1: Write $ 8 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 8 \cdot \frac{x^3}{x^2+4x} & \xlongequal{\text{Step 1}} \frac{8}{\color{red}{1}} \cdot \frac{x^3}{x^2+4x} \xlongequal{\text{Step 2}} \frac{ 8 \cdot x^3 }{ 1 \cdot \left( x^2+4x \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 8x^3 }{ x^2+4x } \end{aligned} $$ |
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Simplify Rational Expressions