Answer
$$(-100x^2+900x-1800)\cdot\frac{2}{3}(x-3)=\frac{-200x^3+2400x^2-9000x+10800}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(-100x^2+900x-1800)\cdot\frac{2}{3}(x-3)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-200x^2+1800x-3600}{3}(x-3) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-200x^3+2400x^2-9000x+10800}{3}\end{aligned} $$ | |
| ① | Multiply $-100x^2+900x-1800$ by $ \dfrac{2}{3} $ to get $ \dfrac{ -200x^2+1800x-3600 }{ 3 } $. Step 1: Write $ -100x^2+900x-1800 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} -100x^2+900x-1800 \cdot \frac{2}{3} & \xlongequal{\text{Step 1}} \frac{-100x^2+900x-1800}{\color{red}{1}} \cdot \frac{2}{3} \xlongequal{\text{Step 2}} \frac{ \left( -100x^2+900x-1800 \right) \cdot 2 }{ 1 \cdot 3 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -200x^2+1800x-3600 }{ 3 } \end{aligned} $$ |
| ② | Multiply $ \dfrac{-200x^2+1800x-3600}{3} $ by $ x-3 $ to get $ \dfrac{-200x^3+2400x^2-9000x+10800}{3} $. Step 1: Write $ x-3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{-200x^2+1800x-3600}{3} \cdot x-3 & \xlongequal{\text{Step 1}} \frac{-200x^2+1800x-3600}{3} \cdot \frac{x-3}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( -200x^2+1800x-3600 \right) \cdot \left( x-3 \right) }{ 3 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ -200x^3+600x^2+1800x^2-5400x-3600x+10800 }{ 3 } = \\[1ex] &= \frac{-200x^3+2400x^2-9000x+10800}{3} \end{aligned} $$ |
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