Multiply each term of $ \left( \color{blue}{x+2}\right) $ by each term in $ \left( x+1\right) $.

$$ \left( \color{blue}{x+2}\right) \cdot \left( x+1\right) = x^2+x+2x+2 $$

Combine like terms:

$$ x^2+ \color{blue}{x} + \color{blue}{2x} +2 = x^2+ \color{blue}{3x} +2 $$

Combine like terms:

$$ \color{blue}{x} + \color{red}{2} +x^2+ \color{blue}{3x} + \color{red}{2} = x^2+ \color{blue}{4x} + \color{red}{4} $$

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 }$.

$$ \begin{aligned}\left(x+1\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 1 + \color{red}{1^2} = x^2+2x+1\end{aligned} $$

Find $ \left(x+1\right)^3 $ using formula

$$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$

where $ A = x $ and $ B = 1 $.

$$ \left(x+1\right)^3 = x^3+3 \cdot x^2 \cdot 1 + 3 \cdot x \cdot 1^2+1^3 = x^3+3x^2+3x+1 $$

Multiply $x+2$ by $ \dfrac{x^2+2x+1}{2} $ to get $ \dfrac{x^3+4x^2+5x+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.

$$ \begin{aligned} x+2 \cdot \frac{x^2+2x+1}{2} & \xlongequal{\text{Step 1}} \frac{x+2}{\color{red}{1}} \cdot \frac{x^2+2x+1}{2} \xlongequal{\text{Step 2}} \frac{ \left( x+2 \right) \cdot \left( x^2+2x+1 \right) }{ 1 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^3+2x^2+x+2x^2+4x+2 }{ 2 } = \frac{x^3+4x^2+5x+2}{2} \end{aligned} $$

Multiply $x+2$ by $ \dfrac{x^3+3x^2+3x+1}{3} $ to get $ \dfrac{x^4+5x^3+9x^2+7x+2}{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.

$$ \begin{aligned} x+2 \cdot \frac{x^3+3x^2+3x+1}{3} & \xlongequal{\text{Step 1}} \frac{x+2}{\color{red}{1}} \cdot \frac{x^3+3x^2+3x+1}{3} \xlongequal{\text{Step 2}} \frac{ \left( x+2 \right) \cdot \left( x^3+3x^2+3x+1 \right) }{ 1 \cdot 3 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^4+3x^3+3x^2+x+2x^3+6x^2+6x+2 }{ 3 } = \frac{x^4+5x^3+9x^2+7x+2}{3} \end{aligned} $$

Add $x^2+4x+4$ and $ \dfrac{x^3+4x^2+5x+2}{2} $ to get $ \dfrac{ \color{purple}{ x^3+6x^2+13x+10 } }{ 2 }$.

Step 1: Write $ x^2+4x+4 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 2 }$.

$$ \begin{aligned} x^2+4x+4+ \frac{x^3+4x^2+5x+2}{2} & \xlongequal{\text{Step 1}} \frac{x^2+4x+4}{\color{red}{1}} + \frac{x^3+4x^2+5x+2}{2} = \frac{ \left( x^2+4x+4 \right) \cdot \color{blue}{ 2 }}{ 1 \cdot \color{blue}{ 2 }} + \frac{ x^3+4x^2+5x+2 }{ 2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x^2+8x+8 } }{ 2 } + \frac{ \color{purple}{ x^3+4x^2+5x+2 } }{ 2 }=\frac{ \color{purple}{ x^3+6x^2+13x+10 } }{ 2 } \end{aligned} $$

Multiply $x+2$ by $ \dfrac{x^3+3x^2+3x+1}{3} $ to get $ \dfrac{x^4+5x^3+9x^2+7x+2}{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.

$$ \begin{aligned} x+2 \cdot \frac{x^3+3x^2+3x+1}{3} & \xlongequal{\text{Step 1}} \frac{x+2}{\color{red}{1}} \cdot \frac{x^3+3x^2+3x+1}{3} \xlongequal{\text{Step 2}} \frac{ \left( x+2 \right) \cdot \left( x^3+3x^2+3x+1 \right) }{ 1 \cdot 3 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^4+3x^3+3x^2+x+2x^3+6x^2+6x+2 }{ 3 } = \frac{x^4+5x^3+9x^2+7x+2}{3} \end{aligned} $$

Add $ \dfrac{x^3+6x^2+13x+10}{2} $ and $ \dfrac{x^4+5x^3+9x^2+7x+2}{3} $ to get $ \dfrac{ \color{purple}{ 2x^4+13x^3+36x^2+53x+34 } }{ 6 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ 3 }$ and the second by $\color{blue}{ 2 }$.

$$ \begin{aligned} \frac{x^3+6x^2+13x+10}{2} + \frac{x^4+5x^3+9x^2+7x+2}{3} & = \frac{ \left( x^3+6x^2+13x+10 \right) \cdot \color{blue}{ 3 }}{ 2 \cdot \color{blue}{ 3 }} + \frac{ \left( x^4+5x^3+9x^2+7x+2 \right) \cdot \color{blue}{ 2 }}{ 3 \cdot \color{blue}{ 2 }} = \\[1ex] &=\frac{ \color{purple}{ 3x^3+18x^2+39x+30 } }{ 6 } + \frac{ \color{purple}{ 2x^4+10x^3+18x^2+14x+4 } }{ 6 } = \\[1ex] &=\frac{ \color{purple}{ 2x^4+13x^3+36x^2+53x+34 } }{ 6 } \end{aligned} $$