Find $ \left(2k+3\right)^2 $ using formula.

$$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ 2k } $ and $ B = \color{red}{ 3 }$.

$$ \begin{aligned}\left(2k+3\right)^2 = \color{blue}{\left( 2k \right)^2} +2 \cdot 2k \cdot 3 + \color{red}{3^2} = 4k^2+12k+9\end{aligned} $$

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

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

Combine like terms:

$$ 2k^2+ \color{blue}{k} + \color{blue}{2k} +1 = 2k^2+ \color{blue}{3k} +1 $$

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

$$ \left( \color{blue}{2k^2+3k+1}\right) \cdot \left( 2k+3\right) = 4k^3+6k^2+6k^2+9k+2k+3 $$

Combine like terms:

$$ 4k^3+ \color{blue}{6k^2} + \color{blue}{6k^2} + \color{red}{9k} + \color{red}{2k} +3 = 4k^3+ \color{blue}{12k^2} + \color{red}{11k} +3 $$

Add $ \dfrac{4k^3+12k^2+11k+3}{3} $ and $ 4k^2+12k+9 $ to get $ \dfrac{ \color{purple}{ 4k^3+24k^2+47k+30 } }{ 3 }$.

Step 1: Write $ 4k^2+12k+9 $ 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 second fraction by $\color{blue}{3}$.

$$ \begin{aligned} \frac{4k^3+12k^2+11k+3}{3} +4k^2+12k+9 & \xlongequal{\text{Step 1}} \frac{4k^3+12k^2+11k+3}{3} + \frac{4k^2+12k+9}{\color{red}{1}} = \\[1ex] &= \frac{ 4k^3+12k^2+11k+3 }{ 3 } + \frac{ \left( 4k^2+12k+9 \right) \cdot \color{blue}{ 3 }}{ 1 \cdot \color{blue}{ 3 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 4k^3+12k^2+11k+3 } }{ 3 } + \frac{ \color{purple}{ 12k^2+36k+27 } }{ 3 }=\frac{ \color{purple}{ 4k^3+24k^2+47k+30 } }{ 3 } \end{aligned} $$