Math Problem Statement

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The area of a rectangle is 8x^2 - 32. The length of the rectangle is x - 2. Which of the following expressions could represent the width of the rectangle?

Solution

We are given that the area of the rectangle is 8x2328x^2 - 32 and the length is x2x - 2. The task is to determine which of the provided expressions could represent the width of the rectangle.

Step 1: Area Formula for a Rectangle

The area of a rectangle is calculated using the formula: Area=Length×Width\text{Area} = \text{Length} \times \text{Width} Here, the given area is 8x2328x^2 - 32, and the length is x2x - 2. Let the width be WW. So: Area=(Length)×(Width)\text{Area} = (\text{Length}) \times (\text{Width}) This gives: 8x232=(x2)×W8x^2 - 32 = (x - 2) \times W

Step 2: Solving for the Width

To find WW, divide both sides of the equation by x2x - 2: W=8x232x2W = \frac{8x^2 - 32}{x - 2}

Now, simplify the expression for the width.

Step 3: Factor the Numerator

We can factor the numerator 8x2328x^2 - 32 as follows: 8x232=8(x24)8x^2 - 32 = 8(x^2 - 4) Notice that x24x^2 - 4 is a difference of squares, so it can be factored further: x24=(x2)(x+2)x^2 - 4 = (x - 2)(x + 2) Thus, the numerator becomes: 8(x2)(x+2)8(x - 2)(x + 2)

Step 4: Simplify the Expression

Now, substitute this factored form into the equation for the width: W=8(x2)(x+2)x2W = \frac{8(x - 2)(x + 2)}{x - 2} Since x2x - 2 appears in both the numerator and denominator, we can cancel it out, assuming x2x \neq 2: W=8(x+2)W = 8(x + 2)

Step 5: Identify the Correct Option

We now see that the width of the rectangle is 8(x+2)8(x + 2). However, none of the given options is exactly 8(x+2)8(x + 2), but we can factor out a common factor of 4 from the width expression: W=4×2(x+2)=4(x+2)W = 4 \times 2(x + 2) = 4(x + 2) This matches option A.

Final Answer:

The correct answer is A. 4(x+2)4(x + 2).

Would you like more details or have any questions about this explanation?

Here are 5 related questions for further practice:

  1. If the width of the rectangle was given as 4(x+3)4(x + 3), what would be the area?
  2. What is the geometric significance of factoring the quadratic expression in the area formula?
  3. How does the distributive property help in simplifying expressions in geometry problems?
  4. What would happen if the given expression for the area was not factorizable?
  5. How would the problem change if the length of the rectangle were a different linear expression?

Tip: Always check for common factors in algebraic expressions before attempting more complicated operations like division. It simplifies the problem!

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Math Problem Analysis

Mathematical Concepts

Algebra
Factoring Quadratics
Geometry (Area of Rectangle)

Formulas

Area of Rectangle = Length × Width
Difference of Squares: a^2 - b^2 = (a - b)(a + b)

Theorems

Distributive Property
Factoring of Quadratic Expressions

Suitable Grade Level

Grades 8-10