Problem 75 - College Algebra and Calculus: An Applied Approach

Question

Examination Room A rectangular examination room in a veterinary clinic has a volume of \(x^{3}+11 x^{2}+34 x+24\) cubic feet. The height of the room is \(x+1\) feet (see figure). Find the number of square feet of floor space in the examination room.

Step-by-Step Solution

Answer

The floor space of the room is \(x^{2}+10x+24\) square feet.

Step 1: Understand The Given

We know that the volume of a rectangular prism, or in this case the room, is given by the formula volume = length * width * height. Here, the volume is \(x^{3}+11 x^{2}+34 x+24\) cubic feet and height is \(x+1\) feet.

Step 2: Find the Equation

As we need to find the floor space, we are essentially looking for the area of the base of the room. This base area is given by the formula for volume divided by height. Hence, the floor area is calculated as \(\frac{Volume}{Height}\). In case of our expression this translates to \(\frac{x^{3}+11 x^{2}+34 x+24}{x+1}\)

Step 3: Divide Polynomial by Monomial

Now, divide the given polynomial by the monomial to find the area of the floor (which is a polynomial). This can be done via polynomial division or synthetic division to get floor area as \(x^{2}+10x+24\).

Step 4: Final Answer

So, the floor space of the examination room is given by the polynomial \(x^{2}+10x+24\) square feet.

Key Concepts

Volume of a Rectangular Prism

The volume of a rectangular prism is a foundational concept in geometry. It refers to the amount of space inside the 3-dimensional shape and is calculated by multiplying the prism's length, width, and height together. If you picture a typical shoe box, its volume is determined by how much "stuff" you can fit inside it.
In mathematical terms, the formula for volume is:

• Volume = length × width × height

In the given exercise, the volume of the examination room is represented by the polynomial expression \(x^3 + 11x^2 + 34x + 24\) cubic feet, and the height is given as \(x+1\) feet.
This allows us to understand that the remainder of the expression \((length \times width)\)can be obtained using the volume formula once the height is divided out.
This conceptual understanding is crucial when you're tasked with finding the floor area in problems like these.

Floor Area Calculation

When you want to calculate the floor area of a rectangular prism, or any rectangular room, you're looking to find the size of the flat surface at the base. Essentially, you need the length and width.
In contexts involving polynomial expressions for volume, like in our problem, the floor area is essentially the result of dividing the volume by the height.
This is represented by the formula:

• Floor Area = \(\frac{Volume}{Height}\)

For our examination room, this becomes\(\frac{x^3 + 11x^2 + 34x + 24}{x+1}\).
When you perform polynomial division on these expressions, you find the floor area in terms of another polynomial, which in this case, results in \(x^2 + 10x + 24\) square feet.
Recognizing how these calculations apply builds a strong understanding of geometry and algebra working hand in hand.

Synthetic Division

Synthetic division is a handy mathematical tool used to divide polynomials more quickly than traditional division methods, especially when dealing with linear divisors. It's like a simplified version of long division specifically designed for polynomials.
Without diving into too much complexity, synthetic division is specifically useful when you know you're dividing by a binomial of the form \((x-c)\). However, synthetic division can also be cleverly applied to situations like ours when dividing by linear factors.
In the exercise, when dividing \(x^3 + 11x^2 + 34x + 24\) by \(x+1\) to find the floor area, synthetic division provides a streamlined and efficient approach, reducing potential for error. The result of this division is another polynomial, \(x^2 + 10x + 24\), which represents the sought-after floor area.
The beauty of synthetic division is that it breaks down complex algebraic calculations into easier steps, making it an invaluable technique in algebra.