Answer
$$\frac{27v^2+9v}{27}v=\frac{27v^3+9v^2}{27}$$
Explanation
| $$ \begin{aligned}\frac{27v^2+9v}{27}v& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{27v^3+9v^2}{27}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{27v^2+9v}{27} $ by $ v $ to get $ \dfrac{ 27v^3+9v^2 }{ 27 } $. Step 1: Write $ v $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{27v^2+9v}{27} \cdot v & \xlongequal{\text{Step 1}} \frac{27v^2+9v}{27} \cdot \frac{v}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 27v^2+9v \right) \cdot v }{ 27 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 27v^3+9v^2 }{ 27 } \end{aligned} $$ |
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Simplify Rational Expressions