Question
Answer
$$2{f^4}^2 \cdot \dfrac{f^3}{6}f^9=\dfrac{2f^{20}}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2{f^4}^2\frac{f^3}{6}f^9& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2\cdot1f^8\frac{f^3}{6}f^9 \xlongequal{ } \\[1 em] & \xlongequal{ }2f^8\frac{f^3}{6}f^9 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2f^{11}}{6}f^9 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2f^{20}}{6}\end{aligned} $$ | |
| ① | $$ \left( f^4 \right)^2 = 1^2 \left( f^4 \right)^2 = f^8 $$ |
| ② | Multiply $2f^8$ by $ \dfrac{f^3}{6} $ to get $ \dfrac{ 2f^{11} }{ 6 } $. Step 1: Write $ 2f^8 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 2f^8 \cdot \frac{f^3}{6} & \xlongequal{\text{Step 1}} \frac{2f^8}{\color{red}{1}} \cdot \frac{f^3}{6} \xlongequal{\text{Step 2}} \frac{ 2f^8 \cdot f^3 }{ 1 \cdot 6 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2f^{11} }{ 6 } \end{aligned} $$ |
| ③ | Multiply $ \dfrac{2f^{11}}{6} $ by $ f^9 $ to get $ \dfrac{ 2f^{20} }{ 6 } $. Step 1: Write $ f^9 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{2f^{11}}{6} \cdot f^9 & \xlongequal{\text{Step 1}} \frac{2f^{11}}{6} \cdot \frac{f^9}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 2f^{11} \cdot f^9 }{ 6 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2f^{20} }{ 6 } \end{aligned} $$ |
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