Problem 368 - Precalculus

Question

Use the given volume of a box and its length and width to express the height of the box algebraically. Volume is \(12 x^{3}+20 x^{2}-21 x-36,\) length is \(2 x+3,\) width is \(3 x-4.\)

Step-by-Step Solution

Answer

Height, \(h(x) = \frac{12x^{3}+20x^{2}-21x-36}{6x^2 + x - 12}\).

Step 1: Understand the Task

We are given the volume of the box and need to express the height algebraically given the length and width. The formula for the volume of a box is\[\text{Volume} = \text{Length} \times \text{Width} \times \text{Height}.\]

Step 2: Identify Known Values

The given polynomial for the volume is \(12x^{3}+20x^{2}-21x-36\). The length of the box is \(2x+3\) and the width is \(3x-4\). We need to express the height \(h(x)\) in terms of \(x\).

Step 3: Setup the Equation for Volume

Write the equation for volume as \[12x^{3}+20x^{2}-21x-36 = (2x+3)(3x-4)h(x).\]

Step 4: Simplify the Product of Length and Width

Multiply the two binomials \((2x+3)(3x-4)\) to find their product: \[(2x+3)(3x-4) = 6x^2 - 8x + 9x - 12 = 6x^2 + x - 12.\]

Step 5: Solve for Height

Substitute the product of length and width back into the volume equation and solve for \(h(x)\): \[12x^{3}+20x^{2}-21x-36 = (6x^2 + x - 12)h(x).\] Then rearrange to find \[h(x) = \frac{12x^{3}+20x^{2}-21x-36}{6x^2 + x - 12}.\]

Step 6: Confirm Solution

The expression \(h(x) = \frac{12x^{3}+20x^{2}-21x-36}{6x^2 + x - 12}\) represents the height algebraically. This fraction represents the height as it provides a simplified form once the numerator is divided by the polynomial expressed in the denominator.

Key Concepts

Volume of a Box

Understanding the volume of a box is key to tackling problems related to dimensions. The volume represents the amount of space inside the box. You can find it by multiplying the box's length, width, and height. The formula used is:\[ \text{Volume} = \text{Length} \times \text{Width} \times \text{Height} \]This formula is fundamental in many practical applications and helps in visualizing how changes in any dimension affect the overall space within the box.

• Length: The longest side of the box and one part of the volume calculation.
• Width: Perpendicular to the length, contributing to the base area of the box.
• Height: The final dimension, dictating how tall the enclosure is.

By knowing any two of these elements and the volume, you can algebraically determine the missing dimension, such as height.

Multiplying Binomials

To find the product of binomials, like the length and width in this problem, you apply the distributive property, often called FOIL when dealing with two binomials. This acronym helps remember the steps: First, Outside, Inside, Last. Let's see this in action with:\((2x + 3)(3x - 4)\).

• First: Multiply the first terms: \(2x \times 3x = 6x^2\).
• Outside: Multiply the outer terms: \(2x \times -4 = -8x\).
• Inside: Multiply the inner terms: \(3 \times 3x = 9x\).
• Last: Multiply the last terms: \(3 \times -4 = -12\).

Adding all these together gives the expanded form: \(6x^2 - 8x + 9x - 12 = 6x^2 + x - 12\). This resulting polynomial is the combined result of the length and width, which is crucial for determining the height from the volume.

Algebraic Expressions

Algebraic expressions consist of variables, constants, and arithmetic operations. They allow us to represent real-world quantities in mathematical form. An essential skill in algebra is simplifying these expressions, often using operations like addition, subtraction, multiplication, and division. In this exercise, the goal was to express the height of the box as a simplified algebraic expression. You start by using the polynomial for volume and factor it into known elements, i.e., the length and width. This involves dividing one polynomial by another to isolate the unknown element, which in our problem is the height:\[h(x) = \frac{12x^{3}+20x^{2}-21x-36}{6x^2 + x - 12}\]This fraction allows you to express the function of height in terms of \(x\), providing a complete understanding of the problem and enabling further calculations or evaluations if necessary. Recognizing how to transform complex expressions into more manageable forms is vital in algebra.