$$\dfrac{9}{10}(x+3)x(x+1)(x-1)=\dfrac{9x^4+27x^3-9x^2-27x}{10}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{9}{10}(x+3)x(x+1)(x-1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{9x+27}{10}x(x+1)(x-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{9x^2+27x}{10}(x+1)(x-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{9x^3+36x^2+27x}{10}(x-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{9x^4+27x^3-9x^2-27x}{10}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{9}{10} $ by $ x+3 $ to get $ \dfrac{ 9x+27 }{ 10 } $. 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. $$ \begin{aligned} \frac{9}{10} \cdot x+3 & \xlongequal{\text{Step 1}} \frac{9}{10} \cdot \frac{x+3}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 9 \cdot \left( x+3 \right) }{ 10 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 9x+27 }{ 10 } \end{aligned} $$ |
| ② | Multiply $ \dfrac{9x+27}{10} $ by $ x $ to get $ \dfrac{ 9x^2+27x }{ 10 } $. 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. $$ \begin{aligned} \frac{9x+27}{10} \cdot x & \xlongequal{\text{Step 1}} \frac{9x+27}{10} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 9x+27 \right) \cdot x }{ 10 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 9x^2+27x }{ 10 } \end{aligned} $$ |
| ③ | Multiply $ \dfrac{9x^2+27x}{10} $ by $ x+1 $ to get $ \dfrac{9x^3+36x^2+27x}{10} $. Step 1: Write $ x+1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{9x^2+27x}{10} \cdot x+1 & \xlongequal{\text{Step 1}} \frac{9x^2+27x}{10} \cdot \frac{x+1}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 9x^2+27x \right) \cdot \left( x+1 \right) }{ 10 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 9x^3+9x^2+27x^2+27x }{ 10 } = \frac{9x^3+36x^2+27x}{10} \end{aligned} $$ |
| ④ | Multiply $ \dfrac{9x^3+36x^2+27x}{10} $ by $ x-1 $ to get $ \dfrac{9x^4+27x^3-9x^2-27x}{10} $. Step 1: Write $ x-1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{9x^3+36x^2+27x}{10} \cdot x-1 & \xlongequal{\text{Step 1}} \frac{9x^3+36x^2+27x}{10} \cdot \frac{x-1}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 9x^3+36x^2+27x \right) \cdot \left( x-1 \right) }{ 10 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 9x^4-9x^3+36x^3-36x^2+27x^2-27x }{ 10 } = \frac{9x^4+27x^3-9x^2-27x}{10} \end{aligned} $$ |