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