Multiply each term of $ \left( \color{blue}{x+4}\right) $ by each term in $ \left( 2x-1\right) $.

$$ \left( \color{blue}{x+4}\right) \cdot \left( 2x-1\right) = 2x^2-x+8x-4 $$

Combine like terms:

$$ 2x^2 \color{blue}{-x} + \color{blue}{8x} -4 = 2x^2+ \color{blue}{7x} -4 $$

Multiply each term of $ \left( \color{blue}{x+8}\right) $ by each term in $ \left( 2x-1\right) $.

$$ \left( \color{blue}{x+8}\right) \cdot \left( 2x-1\right) = 2x^2-x+16x-8 $$

Combine like terms:

$$ 2x^2 \color{blue}{-x} + \color{blue}{16x} -8 = 2x^2+ \color{blue}{15x} -8 $$

Multiply each term of $ \left( \color{blue}{2x^2+15x-8}\right) $ by each term in $ \left( 3x-4\right) $.

$$ \left( \color{blue}{2x^2+15x-8}\right) \cdot \left( 3x-4\right) = 6x^3-8x^2+45x^2-60x-24x+32 $$

Combine like terms:

$$ 6x^3 \color{blue}{-8x^2} + \color{blue}{45x^2} \color{red}{-60x} \color{red}{-24x} +32 = 6x^3+ \color{blue}{37x^2} \color{red}{-84x} +32 $$

Multiply $2x^2+7x-4$ by $ \dfrac{x-7}{6x^3+37x^2-84x+32} $ to get $ \dfrac{x^2-3x-28}{3x^2+20x-32} $.

Step 1: Write $ 2x^2+7x-4 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Factor numerators and denominators.

Step 3: Cancel common factors.

Step 4: Multiply numerators and denominators.

Step 5: Simplify numerator and denominator.

$$ \begin{aligned} 2x^2+7x-4 \cdot \frac{x-7}{6x^3+37x^2-84x+32} & \xlongequal{\text{Step 1}} \frac{2x^2+7x-4}{\color{red}{1}} \cdot \frac{x-7}{6x^3+37x^2-84x+32} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( x+4 \right) \cdot \color{blue}{ \left( 2x-1 \right) } }{ 1 } \cdot \frac{ x-7 }{ \left( 3x^2+20x-32 \right) \cdot \color{blue}{ \left( 2x-1 \right) } } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x+4 }{ 1 } \cdot \frac{ x-7 }{ 3x^2+20x-32 } \xlongequal{\text{Step 4}} \frac{ \left( x+4 \right) \cdot \left( x-7 \right) }{ 1 \cdot \left( 3x^2+20x-32 \right) } = \\[1ex] & \xlongequal{\text{Step 5}} \frac{ x^2-7x+4x-28 }{ 3x^2+20x-32 } = \frac{x^2-3x-28}{3x^2+20x-32} \end{aligned} $$