Multiply $ \color{blue}{12} $ by $ \left( 2t+1\right) $ $$ \color{blue}{12} \cdot \left( 2t+1\right) = 24t+12 $$

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

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

Multiply $2$ by $ \dfrac{2t^3+9t^2+28t+84}{24t+12} $ to get $ \dfrac{ 4t^3+18t^2+56t+168 }{ 24t+12 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 2 \cdot \frac{2t^3+9t^2+28t+84}{24t+12} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{2t^3+9t^2+28t+84}{24t+12} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 2 \cdot \left( 2t^3+9t^2+28t+84 \right) }{ 1 \cdot \left( 24t+12 \right) } \xlongequal{\text{Step 3}} \frac{ 4t^3+18t^2+56t+168 }{ 24t+12 } \end{aligned} $$

Multiply each term of $ \left( \color{blue}{t^2+6t+8}\right) $ by each term in $ \left( t^2+6t+8\right) $.

$$ \left( \color{blue}{t^2+6t+8}\right) \cdot \left( t^2+6t+8\right) = t^4+6t^3+8t^2+6t^3+36t^2+48t+8t^2+48t+64 $$

Combine like terms:

$$ t^4+ \color{blue}{6t^3} + \color{red}{8t^2} + \color{blue}{6t^3} + \color{green}{36t^2} + \color{orange}{48t} + \color{green}{8t^2} + \color{orange}{48t} +64 = \\ = t^4+ \color{blue}{12t^3} + \color{green}{52t^2} + \color{orange}{96t} +64 $$

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

Subtract $ \dfrac{t^4+12t^3+52t^2+96t+64}{16t^2+16t+4} $ from $ \dfrac{2t^3+9t^2+28t+84}{12t+6} $ to get $ \dfrac{ \color{purple}{ 5t^4+4t^3-26t^2+104t-24 } }{ 48t^2+48t+12 }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ 4t+2 }$ and the second by $\color{blue}{ 3 }$.

$$ \begin{aligned} \frac{2t^3+9t^2+28t+84}{12t+6} - \frac{t^4+12t^3+52t^2+96t+64}{16t^2+16t+4} & = \frac{ \left( 2t^3+9t^2+28t+84 \right) \cdot \color{blue}{ \left( 4t+2 \right) }}{ \left( 12t+6 \right) \cdot \color{blue}{ \left( 4t+2 \right) }} - \frac{ \left( t^4+12t^3+52t^2+96t+64 \right) \cdot \color{blue}{ 3 }}{ \left( 16t^2+16t+4 \right) \cdot \color{blue}{ 3 }} = \\[1ex] &=\frac{ \color{purple}{ 8t^4+4t^3+36t^3+18t^2+112t^2+56t+336t+168 } }{ 48t^2+24t+24t+12 } - \frac{ \color{purple}{ 3t^4+36t^3+156t^2+288t+192 } }{ 48t^2+24t+24t+12 } = \\[1ex] &=\frac{ \color{purple}{ 8t^4+4t^3+36t^3+18t^2+112t^2+56t+336t+168 - \left( 3t^4+36t^3+156t^2+288t+192 \right) } }{ 48t^2+48t+12 } = \\[1ex] &=\frac{ \color{purple}{ 5t^4+4t^3-26t^2+104t-24 } }{ 48t^2+48t+12 } \end{aligned} $$