To solve this problem, let's go through the information provided step-by-step.

Given Information:

• The area of a rectangle is $1,056$ square inches.
• The length of the rectangle is $4$ inches more than $2$ times its width $w$.

Formulating the Equation:

1. Define the Length in Terms of Width: Since the length $L$ is $4$ inches longer than $2$ times the width $w$, we can write: $L = 2w + 4$

2. Area Formula: The area $A$ of a rectangle is given by the formula: $A = \text{length} \times \text{width}$ Substituting the values we have: $1,056 = (2w + 4) \times w$

3. Answer Choice: Among the options provided, the correct equation to use is: $1,056 = (2w + 4) \times w$ This matches the first option: $1,056 = (4 + 2w)w$.

Solving for $w$ (Width of the Rectangle):

1. Expand the equation: $1,056 = 2w^2 + 4w$

2. Rearrange the equation into standard quadratic form: $2w^2 + 4w - 1,056 = 0$

3. Divide by 2 to simplify: $w^2 + 2w - 528 = 0$

4. Solve this quadratic equation using the quadratic formula: $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 2$, and $c = -528$.

5. Calculate: $w = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-528)}}{2 \cdot 1}$ $w = \frac{-2 \pm \sqrt{4 + 2112}}{2}$ $w = \frac{-2 \pm \sqrt{2116}}{2}$ $w = \frac{-2 \pm 46}{2}$

6. This gives two solutions: $w = \frac{44}{2} = 22 \quad \text{or} \quad w = \frac{-48}{2} = -24$ Since width cannot be negative, $w = 22$ inches.

Final Answer:

The width of the rectangle is 22 inches.

Would you like further details, or do you have any questions?

Here are some related questions you may find helpful:

1. How do you use the quadratic formula to solve other area-related problems?
2. What are other ways to solve quadratic equations besides the quadratic formula?
3. How does changing the dimensions of a rectangle affect its area?
4. What happens if you increase both the length and width by the same factor?
5. How can we apply these concepts to solve real-world problems?

Tip: When faced with quadratic equations, always look for simplifications first, like dividing the entire equation by common factors. This makes calculations easier.