Problem 45 - Algebra 2

Question

GEOMETRY The area of rectangle \(A B C D\) is \(6 x^{2}+38 x+56\) square units. Its width is \(2 x+8\) units. What is the length of the rectangle? (lesson 6 - 3 )

Step-by-Step Solution

Answer

The length of the rectangle is \(3x + 7\) units.

Step 1: Recall the Area Formula for a Rectangle

The area of a rectangle is given by the formula \( \text{Area} = \text{Length} \times \text{Width} \). In this problem, we have the area and the width, and we need to find the length.

Step 2: Set Up the Equation

Given that the area is \(6x^2 + 38x + 56\) and the width is \(2x + 8\), we can set up the equation \(6x^2 + 38x + 56 = \text{Length} \times (2x + 8)\).

Step 3: Solve for the Length

To find \(\text{Length}\), divide both sides of the equation by the width \((2x + 8)\): \( \text{Length} = \frac{6x^2 + 38x + 56}{2x + 8}\).

Step 4: Perform Polynomial Division

Perform polynomial division of \(6x^2 + 38x + 56\) by \(2x + 8\):1\. Divide the leading term: \( \frac{6x^2}{2x} = 3x \).2\. Multiply \((2x + 8)\) by \(3x\) and subtract from the original polynomial: \( 6x^2 + 38x + 56 - (6x^2 + 24x) = 14x + 56 \).3\. Divide: \( \frac{14x}{2x} = 7 \).4\. Multiply \((2x + 8)\) by \(7\) and subtract: \( 14x + 56 - (14x + 56) = 0 \).Therefore, the result of the division is \(3x + 7\).

Step 5: Verify the Solution

Multiply the length \((3x + 7)\) by the width \((2x + 8)\) to verify the original area:\[(3x + 7)(2x + 8) = 6x^2 + 38x + 56\], which matches the given area. Thus, the calculation is verified.

Key Concepts

Rectangle Area

Understanding how to calculate the area of rectangles is essential in geometry. The area is essentially the amount of space contained within the perimeter of the rectangle, measured in square units. When you multiply the rectangle's length by its width, it gives you the area.

In mathematical terms, the area formula is expressed as:

• Area = Length × Width

In the exercise, the area of rectangle ABCD is given as a polynomial, which is an expression involving variables, coefficients, and the operations of addition, subtraction, and multiplication.

• Area = \(6x^2 + 38x + 56\)

Here, you know the area's polynomial expression and the width. Your task is to find the length, which involves rearranging the formula to solve for the missing dimension.

Factoring Polynomials

Factoring polynomials is a key method in algebra, which involves expressing a given polynomial as a product of its factors. These factors can be numbers, variables, or polynomial expressions.

To factor a polynomial such as the area of our rectangle, you apply polynomial division. Think of it as reversing the multiplication process to find what two expressions multiply together to give the original polynomial:

• The area polynomial: \(6x^2 + 38x + 56\)
• The width: \(2x + 8\)

We divide the area by the width to find the length. Here's a simplified glimpse of how polynomial division works in the process:

• Divide the leading term of the dividend by the leading term of the divisor.
• Multiply the divisor by the result and subtract from the original polynomial.
• Continue this process until you reach a remainder of zero or can proceed no further.

Ultimately, this division gives us the length, which in this problem is found to be \(3x + 7\).

Geometry Problem Solving

Geometry problem solving often brings together various mathematical principles and methods. In this exercise, we're integrating geometry and algebra. By connecting the area formula with polynomial expressions, we solve for unknown dimensions.

Here's how you might approach a typical problem like this one:

• Identify known values and what needs to be found.
• Set up equations using geometry formulas, expressed as polynomials if given in algebraic form.
• Use algebraic methods such as polynomial division or factoring to solve for unknowns.
• Verify solutions by plugging them back into the original equation to see if they produce the given value.

This methodical approach not only helps in solving specific problems but also develops a deeper understanding of both geometry and algebra. Mastering these connections is crucial for more advanced topics in mathematics.