$$\dfrac{2}{3}(x-2)(x-3)(x+1)^2=\dfrac{2x^4-6x^3-6x^2+14x+12}{3}$$
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2}{3}(x-2)(x-3)(x+1)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{3}(x-2)(x-3)(x^2+2x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2x-4}{3}(x-3)(x^2+2x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2x^2-10x+12}{3}(x^2+2x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2x^4-6x^3-6x^2+14x+12}{3}\end{aligned} $$ | |
| ① | Find $ \left(x+1\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}{ 1 }$. $$ \begin{aligned}\left(x+1\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 1 + \color{red}{1^2} = x^2+2x+1\end{aligned} $$ |
| ② | Multiply $ \dfrac{2}{3} $ by $ x-2 $ to get $ \dfrac{ 2x-4 }{ 3 } $. 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. $$ \begin{aligned} \frac{2}{3} \cdot x-2 & \xlongequal{\text{Step 1}} \frac{2}{3} \cdot \frac{x-2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 2 \cdot \left( x-2 \right) }{ 3 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x-4 }{ 3 } \end{aligned} $$ |
| ③ | Multiply $ \dfrac{2x-4}{3} $ by $ x-3 $ to get $ \dfrac{2x^2-10x+12}{3} $. Step 1: Write $ x-3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{2x-4}{3} \cdot x-3 & \xlongequal{\text{Step 1}} \frac{2x-4}{3} \cdot \frac{x-3}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 2x-4 \right) \cdot \left( x-3 \right) }{ 3 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x^2-6x-4x+12 }{ 3 } = \frac{2x^2-10x+12}{3} \end{aligned} $$ |
| ④ | Multiply $ \dfrac{2x^2-10x+12}{3} $ by $ x^2+2x+1 $ to get $ \dfrac{2x^4-6x^3-6x^2+14x+12}{3} $. 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. $$ \begin{aligned} \frac{2x^2-10x+12}{3} \cdot x^2+2x+1 & \xlongequal{\text{Step 1}} \frac{2x^2-10x+12}{3} \cdot \frac{x^2+2x+1}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( 2x^2-10x+12 \right) \cdot \left( x^2+2x+1 \right) }{ 3 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 2x^4+4x^3+2x^2-10x^3-20x^2-10x+12x^2+24x+12 }{ 3 } = \\[1ex] &= \frac{2x^4-6x^3-6x^2+14x+12}{3} \end{aligned} $$ |