Find $ \left(x+2\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}{ 2 }$.

$$ \begin{aligned}\left(x+2\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 2 + \color{red}{2^2} = x^2+4x+4\end{aligned} $$

Add $7+x$ and $ \dfrac{5}{x} $ to get $ \dfrac{ \color{purple}{ x^2+7x+5 } }{ x }$.

Step 1: Write $ 7+x $ 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 $\color{blue}{ x }$.

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

Multiply $ \dfrac{x^2+7x+5}{x} $ by $ x^2+4x+4 $ to get $ \dfrac{x^4+11x^3+37x^2+48x+20}{x} $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x^2+7x+5}{x} \cdot x^2+4x+4 & \xlongequal{\text{Step 1}} \frac{x^2+7x+5}{x} \cdot \frac{x^2+4x+4}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( x^2+7x+5 \right) \cdot \left( x^2+4x+4 \right) }{ x \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^4+4x^3+4x^2+7x^3+28x^2+28x+5x^2+20x+20 }{ x } = \frac{x^4+11x^3+37x^2+48x+20}{x} \end{aligned} $$