Question
Answer
$$\dfrac{1}{4}(n^2-8n+15)=\dfrac{n^2-8n+15}{4}$$
Explanation
| $$ \begin{aligned}\frac{1}{4}(n^2-8n+15)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{n^2-8n+15}{4}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{1}{4} $ by $ n^2-8n+15 $ to get $ \dfrac{ n^2-8n+15 }{ 4 } $. Step 1: Write $ n^2-8n+15 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{1}{4} \cdot n^2-8n+15 & \xlongequal{\text{Step 1}} \frac{1}{4} \cdot \frac{n^2-8n+15}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot \left( n^2-8n+15 \right) }{ 4 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ n^2-8n+15 }{ 4 } \end{aligned} $$ |
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