$$\dfrac{1}{4}(x^4-1)-\dfrac{1}{4}(x^2+2x+3)(x^2-2x+1)=\dfrac{4x-4}{4}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{4}(x^4-1)-\frac{1}{4}(x^2+2x+3)(x^2-2x+1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^4-1}{4}-\frac{x^2+2x+3}{4}(x^2-2x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{x^4-1}{4}-\frac{x^4-4x+3}{4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{4x-4}{4}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{1}{4} $ by $ x^4-1 $ to get $ \dfrac{ x^4-1 }{ 4 } $. Step 1: Write $ x^4-1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{1}{4} \cdot x^4-1 & \xlongequal{\text{Step 1}} \frac{1}{4} \cdot \frac{x^4-1}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot \left( x^4-1 \right) }{ 4 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^4-1 }{ 4 } \end{aligned} $$ |
| ② | Multiply $ \dfrac{1}{4} $ by $ x^2+2x+3 $ to get $ \dfrac{ x^2+2x+3 }{ 4 } $. Step 1: Write $ x^2+2x+3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{1}{4} \cdot x^2+2x+3 & \xlongequal{\text{Step 1}} \frac{1}{4} \cdot \frac{x^2+2x+3}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot \left( x^2+2x+3 \right) }{ 4 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^2+2x+3 }{ 4 } \end{aligned} $$ |
| ③ | Multiply $ \dfrac{1}{4} $ by $ x^4-1 $ to get $ \dfrac{ x^4-1 }{ 4 } $. Step 1: Write $ x^4-1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{1}{4} \cdot x^4-1 & \xlongequal{\text{Step 1}} \frac{1}{4} \cdot \frac{x^4-1}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot \left( x^4-1 \right) }{ 4 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^4-1 }{ 4 } \end{aligned} $$ |
| ④ | Multiply $ \dfrac{x^2+2x+3}{4} $ by $ x^2-2x+1 $ to get $ \dfrac{x^4-4x+3}{4} $. Step 1: Write $ x^2-2x+1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{x^2+2x+3}{4} \cdot x^2-2x+1 & \xlongequal{\text{Step 1}} \frac{x^2+2x+3}{4} \cdot \frac{x^2-2x+1}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( x^2+2x+3 \right) \cdot \left( x^2-2x+1 \right) }{ 4 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^4 -\cancel{2x^3}+x^2+ \cancel{2x^3}-4x^2+2x+3x^2-6x+3 }{ 4 } = \frac{x^4-4x+3}{4} \end{aligned} $$ |
| ⑤ | Subtract $ \dfrac{x^4-4x+3}{4} $ from $ \dfrac{x^4-1}{4} $ to get $ \dfrac{4x-4}{4} $. To subtract expressions with the same denominators, we subtract the numerators and write the result over the common denominator. $$ \begin{aligned} \frac{x^4-1}{4} - \frac{x^4-4x+3}{4} & = \frac{x^4-1}{\color{blue}{4}} - \frac{x^4-4x+3}{\color{blue}{4}} = \\[1ex] &=\frac{ x^4-1 - \left( x^4-4x+3 \right) }{ \color{blue}{ 4 }}= \frac{4x-4}{4} \end{aligned} $$ |