Find $ \left(2k+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}{ 2k } $ and $ B = \color{red}{ 1 }$.

$$ \begin{aligned}\left(2k+1\right)^2 = \color{blue}{\left( 2k \right)^2} +2 \cdot 2k \cdot 1 + \color{red}{1^2} = 4k^2+4k+1\end{aligned} $$

Multiply $ \color{blue}{4} $ by $ \left( 2k+1\right) $ $$ \color{blue}{4} \cdot \left( 2k+1\right) = 8k+4 $$Multiply $ \color{blue}{8} $ by $ \left( 2k+1\right) $ $$ \color{blue}{8} \cdot \left( 2k+1\right) = 16k+8 $$

$$ \left( \color{blue}{8k+4}\right) \cdot n = 8kn+4n $$

Combine like terms:

$$ 4n^2+8kn+4n = 8kn+4n^2+4n $$

Combine like terms:

$$ 8kn+4n^2+ \color{blue}{4n} \color{blue}{-16n} = 8kn+4n^2 \color{blue}{-12n} $$

Combine like terms:

$$ 8kn+4n^2-12n+4k^2+4k+1 = 4k^2+8kn+4n^2+4k-12n+1 $$

Remove the parentheses by changing the sign of each term within them.

$$ - \left( 16k+8 \right) = -16k-8 $$

Combine like terms:

$$ 4k^2+8kn+4n^2+ \color{blue}{4k} -12n+ \color{red}{1} \color{blue}{-16k} \color{green}{-8} + \color{green}{15} = 4k^2+8kn+4n^2 \color{blue}{-12k} -12n+ \color{green}{8} $$

Multiply $ \dfrac{1}{8} $ by $ 4k^2+8kn+4n^2-12k-12n+8 $ to get $ \dfrac{ 4k^2+8kn+4n^2-12k-12n+8 }{ 8 } $.

Step 1: Write $ 4k^2+8kn+4n^2-12k-12n+8 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{1}{8} \cdot 4k^2+8kn+4n^2-12k-12n+8 & \xlongequal{\text{Step 1}} \frac{1}{8} \cdot \frac{4k^2+8kn+4n^2-12k-12n+8}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 1 \cdot \left( 4k^2+8kn+4n^2-12k-12n+8 \right) }{ 8 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 4k^2+8kn+4n^2-12k-12n+8 }{ 8 } \end{aligned} $$