$$-\dfrac{1}{2}(x-1)\dfrac{1}{2}(x-2)(x-6)(2x-7)=\dfrac{2x^4-25x^3+103x^2-164x+84}{4}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-\frac{1}{2}(x-1)\frac{1}{2}(x-2)(x-6)(2x-7)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x-1}{2}\frac{x-2}{2}(x-6)(2x-7) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{x^2-3x+2}{4}(x-6)(2x-7) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{x^3-9x^2+20x-12}{4}(2x-7) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{2x^4-25x^3+103x^2-164x+84}{4}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{1}{2} $ by $ x-1 $ to get $ \dfrac{ x-1 }{ 2 } $. 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. $$ \begin{aligned} \frac{1}{2} \cdot x-1 & \xlongequal{\text{Step 1}} \frac{1}{2} \cdot \frac{x-1}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot \left( x-1 \right) }{ 2 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x-1 }{ 2 } \end{aligned} $$ |
| ② | Multiply $ \dfrac{1}{2} $ by $ x-2 $ to get $ \dfrac{ x-2 }{ 2 } $. 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. $$ \begin{aligned} \frac{1}{2} \cdot x-2 & \xlongequal{\text{Step 1}} \frac{1}{2} \cdot \frac{x-2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot \left( x-2 \right) }{ 2 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x-2 }{ 2 } \end{aligned} $$ |
| ③ | Multiply $ \dfrac{x-1}{2} $ by $ \dfrac{x-2}{2} $ to get $ \dfrac{x^2-3x+2}{4} $. Step 1: Multiply numerators and denominators. Step 2: Simplify numerator and denominator. $$ \begin{aligned} \frac{x-1}{2} \cdot \frac{x-2}{2} & \xlongequal{\text{Step 1}} \frac{ \left( x-1 \right) \cdot \left( x-2 \right) }{ 2 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ x^2-2x-x+2 }{ 4 } = \frac{x^2-3x+2}{4} \end{aligned} $$ |
| ④ | Multiply $ \dfrac{x^2-3x+2}{4} $ by $ x-6 $ to get $ \dfrac{x^3-9x^2+20x-12}{4} $. Step 1: Write $ x-6 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{x^2-3x+2}{4} \cdot x-6 & \xlongequal{\text{Step 1}} \frac{x^2-3x+2}{4} \cdot \frac{x-6}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( x^2-3x+2 \right) \cdot \left( x-6 \right) }{ 4 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^3-6x^2-3x^2+18x+2x-12 }{ 4 } = \frac{x^3-9x^2+20x-12}{4} \end{aligned} $$ |
| ⑤ | Multiply $ \dfrac{x^3-9x^2+20x-12}{4} $ by $ 2x-7 $ to get $ \dfrac{2x^4-25x^3+103x^2-164x+84}{4} $. Step 1: Write $ 2x-7 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{x^3-9x^2+20x-12}{4} \cdot 2x-7 & \xlongequal{\text{Step 1}} \frac{x^3-9x^2+20x-12}{4} \cdot \frac{2x-7}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( x^3-9x^2+20x-12 \right) \cdot \left( 2x-7 \right) }{ 4 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 2x^4-7x^3-18x^3+63x^2+40x^2-140x-24x+84 }{ 4 } = \\[1ex] &= \frac{2x^4-25x^3+103x^2-164x+84}{4} \end{aligned} $$ |