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 }$.

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

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 }$.

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

Multiply each term of $ \left( \color{blue}{x^2+2x+1}\right) $ by each term in $ \left( x-3\right) $.

$$ \left( \color{blue}{x^2+2x+1}\right) \cdot \left( x-3\right) = x^3-3x^2+2x^2-6x+x-3 $$

Combine like terms:

$$ x^3 \color{blue}{-3x^2} + \color{blue}{2x^2} \color{red}{-6x} + \color{red}{x} -3 = x^3 \color{blue}{-x^2} \color{red}{-5x} -3 $$

Simplify $ \dfrac{x-3}{x^3-x^2-5x-3} $ to $ \dfrac{1}{x^2+2x+1} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-3}$.

$$ \begin{aligned} \frac{x-3}{x^3-x^2-5x-3} & =\frac{ 1 \cdot \color{blue}{ \left( x-3 \right) }}{ \left( x^2+2x+1 \right) \cdot \color{blue}{ \left( x-3 \right) }} = \\[1ex] &= \frac{1}{x^2+2x+1} \end{aligned} $$

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

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.

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