Problem 76
Question
Veterinary Clinic A rectangular veterinary clinic has a volume of \(x^{3}+55 x^{2}+650 x+2000\) cubic feet (the space in the attic is not counted). The height of the clinic is \(x+5\) feet (see figure). Find the number of square feet of floor space on the first floor of the clinic.
Step-by-Step Solution
Verified Answer
The area of the floor space on the first floor of the veterinary clinic is \(x^{2}+50x+400\) square feet.
1Step 1: Understand the problem
The problem provides the cubic volume and height of a veterinary clinic and asks for the floor space. The volume of a rectangular object is found by multiplying its length, width, and height. We are given that the volume is \(x^{3}+55 x^{2}+650 x+2000\) cubic feet and the height is \(x+5\) feet. We can rearrange the formula for volume to solve for the floor space (length times width).
2Step 2: Represent the floor space
Given that the volume \(V = lwh\), we can represent the floor space as \(V/h\). Thus the area \(A = V/h = (x^{3}+55 x^{2}+650 x+2000) / (x+5)\).
3Step 3: Simplify the expression
To simplify, we will need to use polynomial division to divide \(x^{3}+55 x^{2}+650 x+2000\) by \(x+5\). This results in the polynomial \(x^{2}+50x+400\), which represents the floor area in square feet.
4Step 4: Answer the problem
The floor space on the first floor of the clinic is represented by the polynomial \(x^{2}+50x+400\) square feet.
Key Concepts
Polynomial DivisionVolume of a Rectangular PrismSimplifying Algebraic ExpressionsFinding Area
Polynomial Division
Dividing polynomials is a lot like dividing numbers, but instead of numbers, we work with variables like 'x'. Let's say we have a polynomial, like the one representing the volume of the veterinary clinic \(x^3 + 55x^2 + 650x + 2000\) and we want to divide it by another, smaller polynomial such as \(x + 5\).
We can use long division, just as we would with numbers. Divide the first term of the dividend (\the part we're dividing) by the first term of the divisor (what we're dividing by). You'll write the result on top, multiply the divisor by this result, subtract it from the dividend, and bring down the next term. Repeat this process until all terms are brought down. After dividing \(x^3 + 55x^2 + 650x + 2000\) by \(x + 5\), we get \(x^2 + 50x + 400\), which represents the area or the floor space in square feet. This process is essential to simplify algebraic expressions and to find exact values for variable expressions.
We can use long division, just as we would with numbers. Divide the first term of the dividend (\the part we're dividing) by the first term of the divisor (what we're dividing by). You'll write the result on top, multiply the divisor by this result, subtract it from the dividend, and bring down the next term. Repeat this process until all terms are brought down. After dividing \(x^3 + 55x^2 + 650x + 2000\) by \(x + 5\), we get \(x^2 + 50x + 400\), which represents the area or the floor space in square feet. This process is essential to simplify algebraic expressions and to find exact values for variable expressions.
Volume of a Rectangular Prism
In the context of the clinic, we're dealing with a 3-dimensional space. The mathematical representation of this space is through a volume formula of a rectangular prism, which is \(V = l \times w \times h\), where \(l\) is the length, \(w\) is the width, and \(h\) is the height.
Knowing the volume, which in our case is a polynomial \(x^3 + 55x^2 + 650x + 2000\), and one of the dimensions, the height (\(x + 5\)), allows us to find the product of the other two dimensions, the floor area. The formula is reorganized to \(A = \frac{V}{h}\). In practice, solving for the floor area is a two-step process: first find the volume and then divide it by the height.
Knowing the volume, which in our case is a polynomial \(x^3 + 55x^2 + 650x + 2000\), and one of the dimensions, the height (\(x + 5\)), allows us to find the product of the other two dimensions, the floor area. The formula is reorganized to \(A = \frac{V}{h}\). In practice, solving for the floor area is a two-step process: first find the volume and then divide it by the height.
Simplifying Algebraic Expressions
Simplifying algebraic expressions is about making them easier to understand and work with. In the division of polynomials, once you've divided the larger expression, you're left with a more straightforward expression that describes the same thing. In our example, \(x^2 + 50x + 400\) is much simpler than the original volume expression.
It's essential to simplify correctly. Combine like terms – terms that have the same variables to the same power – and perform any arithmetic you can. Our simplified expression for the clinic's floor space is already as simple as it can get, representing a clear and easy-to-use area formula based on the value of \(x\).
It's essential to simplify correctly. Combine like terms – terms that have the same variables to the same power – and perform any arithmetic you can. Our simplified expression for the clinic's floor space is already as simple as it can get, representing a clear and easy-to-use area formula based on the value of \(x\).
Finding Area
The concept of area is central to understanding many real-world applications, including calculating the space inside a building. Area is the amount of two-dimensional space taken up by an object and is measured in square units. For our clinic problem, finding the area of the floor is necessary to determine how much flooring material might be needed, among other uses.
In algebra, areas can be represented by expressions if the dimensions are not concrete numbers. As with our clinic, the area was \(x^2 + 50x + 400\) square feet, which varies depending on the value of \(x\). To find the area in a specific case, we would simply plug in the value for \(x\) and compute the result.
In algebra, areas can be represented by expressions if the dimensions are not concrete numbers. As with our clinic, the area was \(x^2 + 50x + 400\) square feet, which varies depending on the value of \(x\). To find the area in a specific case, we would simply plug in the value for \(x\) and compute the result.
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