$$\dfrac{1}{50}(x-5)^2(x-2)(x+1)(x+4)=\dfrac{x^5-7x^4-11x^3+127x^2-70x-200}{50}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{50}(x-5)^2(x-2)(x+1)(x+4)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{50}(x^2-10x+25)(x-2)(x+1)(x+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^2-10x+25}{50}(x-2)(x+1)(x+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{x^3-12x^2+45x-50}{50}(x+1)(x+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{x^4-11x^3+33x^2-5x-50}{50}(x+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{x^5-7x^4-11x^3+127x^2-70x-200}{50}\end{aligned} $$ | |
| ① | Find $ \left(x-5\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 5 }$. $$ \begin{aligned}\left(x-5\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 5 + \color{red}{5^2} = x^2-10x+25\end{aligned} $$ |
| ② | Multiply $ \dfrac{1}{50} $ by $ x^2-10x+25 $ to get $ \dfrac{ x^2-10x+25 }{ 50 } $. Step 1: Write $ x^2-10x+25 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{1}{50} \cdot x^2-10x+25 & \xlongequal{\text{Step 1}} \frac{1}{50} \cdot \frac{x^2-10x+25}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot \left( x^2-10x+25 \right) }{ 50 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^2-10x+25 }{ 50 } \end{aligned} $$ |
| ③ | Multiply $ \dfrac{x^2-10x+25}{50} $ by $ x-2 $ to get $ \dfrac{x^3-12x^2+45x-50}{50} $. Step 1: Write $ x-2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{x^2-10x+25}{50} \cdot x-2 & \xlongequal{\text{Step 1}} \frac{x^2-10x+25}{50} \cdot \frac{x-2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( x^2-10x+25 \right) \cdot \left( x-2 \right) }{ 50 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^3-2x^2-10x^2+20x+25x-50 }{ 50 } = \frac{x^3-12x^2+45x-50}{50} \end{aligned} $$ |
| ④ | Multiply $ \dfrac{x^3-12x^2+45x-50}{50} $ by $ x+1 $ to get $ \dfrac{x^4-11x^3+33x^2-5x-50}{50} $. 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. $$ \begin{aligned} \frac{x^3-12x^2+45x-50}{50} \cdot x+1 & \xlongequal{\text{Step 1}} \frac{x^3-12x^2+45x-50}{50} \cdot \frac{x+1}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( x^3-12x^2+45x-50 \right) \cdot \left( x+1 \right) }{ 50 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ x^4+x^3-12x^3-12x^2+45x^2+45x-50x-50 }{ 50 } = \\[1ex] &= \frac{x^4-11x^3+33x^2-5x-50}{50} \end{aligned} $$ |
| ⑤ | Multiply $ \dfrac{x^4-11x^3+33x^2-5x-50}{50} $ by $ x+4 $ to get $ \dfrac{x^5-7x^4-11x^3+127x^2-70x-200}{50} $. 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. $$ \begin{aligned} \frac{x^4-11x^3+33x^2-5x-50}{50} \cdot x+4 & \xlongequal{\text{Step 1}} \frac{x^4-11x^3+33x^2-5x-50}{50} \cdot \frac{x+4}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( x^4-11x^3+33x^2-5x-50 \right) \cdot \left( x+4 \right) }{ 50 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ x^5+4x^4-11x^4-44x^3+33x^3+132x^2-5x^2-20x-50x-200 }{ 50 } = \\[1ex] &= \frac{x^5-7x^4-11x^3+127x^2-70x-200}{50} \end{aligned} $$ |