$$-\dfrac{1}{5}(x-5)(x-5)(x+4)=\dfrac{x^3-6x^2-15x+100}{5}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-\frac{1}{5}(x-5)(x-5)(x+4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x-5}{5}(x-5)(x+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^2-10x+25}{5}(x+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{x^3-6x^2-15x+100}{5}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{1}{5} $ by $ x-5 $ to get $ \dfrac{ x-5 }{ 5 } $. Step 1: Write $ x-5 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{1}{5} \cdot x-5 & \xlongequal{\text{Step 1}} \frac{1}{5} \cdot \frac{x-5}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot \left( x-5 \right) }{ 5 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x-5 }{ 5 } \end{aligned} $$ |
| ② | Multiply $ \dfrac{x-5}{5} $ by $ x-5 $ to get $ \dfrac{x^2-10x+25}{5} $. Step 1: Write $ x-5 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{x-5}{5} \cdot x-5 & \xlongequal{\text{Step 1}} \frac{x-5}{5} \cdot \frac{x-5}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( x-5 \right) \cdot \left( x-5 \right) }{ 5 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^2-5x-5x+25 }{ 5 } = \frac{x^2-10x+25}{5} \end{aligned} $$ |
| ③ | Multiply $ \dfrac{x^2-10x+25}{5} $ by $ x+4 $ to get $ \dfrac{x^3-6x^2-15x+100}{5} $. Step 1: Write $ x+4 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{x^2-10x+25}{5} \cdot x+4 & \xlongequal{\text{Step 1}} \frac{x^2-10x+25}{5} \cdot \frac{x+4}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( x^2-10x+25 \right) \cdot \left( x+4 \right) }{ 5 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^3+4x^2-10x^2-40x+25x+100 }{ 5 } = \frac{x^3-6x^2-15x+100}{5} \end{aligned} $$ |