$$-\dfrac{1}{5}x(x-4)^2(x-7)=\dfrac{x^4-15x^3+72x^2-112x}{5}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-\frac{1}{5}x(x-4)^2(x-7)& \xlongequal{ }-\frac{1}{5}x(x^2-8x+16)(x-7) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x}{5}(x^2-8x+16)(x-7) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^3-8x^2+16x}{5}(x-7) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{x^4-15x^3+72x^2-112x}{5}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{1}{5} $ by $ x $ to get $ \dfrac{ x }{ 5 } $. 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. $$ \begin{aligned} \frac{1}{5} \cdot x & \xlongequal{\text{Step 1}} \frac{1}{5} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot x }{ 5 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ x }{ 5 } \end{aligned} $$ |
| ② | Multiply $ \dfrac{x}{5} $ by $ x^2-8x+16 $ to get $ \dfrac{ x^3-8x^2+16x }{ 5 } $. Step 1: Write $ x^2-8x+16 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{x}{5} \cdot x^2-8x+16 & \xlongequal{\text{Step 1}} \frac{x}{5} \cdot \frac{x^2-8x+16}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ x \cdot \left( x^2-8x+16 \right) }{ 5 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^3-8x^2+16x }{ 5 } \end{aligned} $$ |
| ③ | Multiply $ \dfrac{x^3-8x^2+16x}{5} $ by $ x-7 $ to get $ \dfrac{x^4-15x^3+72x^2-112x}{5} $. Step 1: Write $ x-7 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{x^3-8x^2+16x}{5} \cdot x-7 & \xlongequal{\text{Step 1}} \frac{x^3-8x^2+16x}{5} \cdot \frac{x-7}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( x^3-8x^2+16x \right) \cdot \left( x-7 \right) }{ 5 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^4-7x^3-8x^3+56x^2+16x^2-112x }{ 5 } = \frac{x^4-15x^3+72x^2-112x}{5} \end{aligned} $$ |