Problem 93 - Algebra and Trigonometry

Question

Factor the expression completely. (This type of expression arises in calculus in using the “product rule.”) $$ 3(2 x-1)^{2}(2)(x+3)^{1 / 2}+(2 x-1)^{3}\left(\frac{1}{2}\right)(x+3)^{-1 / 2} $$

Step-by-Step Solution

Answer

The expression factors to \(7(2x-1)^2 (x+3)^{-1/2} (x + \frac{5}{2})\).

Step 1: Expand the Expression

Before we factor, it's crucial to better understand the structure of the given expression. We have:\[ 3(2x-1)^2(2)(x+3)^{1/2} + (2x-1)^3\left(\frac{1}{2}\right)(x+3)^{-1/2} \]Multiply the constants to simplify:\[ 6(2x-1)^2(x+3)^{1/2} + \frac{1}{2}(2x-1)^3(x+3)^{-1/2} \]

Step 2: Identify Common Factors

Observe both terms to find common factors. The common factor between the terms is \((2x-1)^2(x+3)^{-1/2}\). Factor it out from each term:\[ (2x-1)^2(x+3)^{-1/2} \left[ 6(x+3) + \frac{1}{2}(2x-1) \right] \]

Step 3: Simplify the Remaining Expression Inside the Brackets

Simplify the expression within the brackets:First, simplify:\[ 6(x+3) = 6x + 18 \]And simplify:\[ \frac{1}{2}(2x-1) = x - \frac{1}{2} \]So the entire expression inside the brackets becomes:\[ 6x + 18 + x - \frac{1}{2} \]

Step 4: Combine Like Terms

Combine the terms within the brackets:\[ 6x + x = 7x \]\[ 18 - \frac{1}{2} = \frac{36}{2} - \frac{1}{2} = \frac{35}{2} \]Thus, the expression inside the brackets is:\[ 7x + \frac{35}{2} \]

Step 5: Final Expression

Now, our expression is:\[ (2x-1)^2 (x+3)^{-1/2} \left( 7x + \frac{35}{2} \right) \]Further factor the expression inside the brackets by taking out the constant:\[ 7 \left( x + \frac{5}{2} \right) \]

Step 6: Write the Factored Form

Bring everything together in the factored form:\[ 7(2x-1)^2 (x+3)^{-1/2} (x + \frac{5}{2}) \]

Key Concepts

The Product Rule and Its Importance

In differentiated calculus, the "product rule" is vital for handling the derivative of two multiplied functions. It applies when you have a product of two functions and want to find its derivative. This is crucial because, in calculus, many functions are compound, made by multiplying easier functions together. Understanding how to differentiate them efficiently is essential.

The mantra of the product rule is: if you have two functions, say \( f(x) \) and \( g(x) \), the derivative of their product \( f(x)g(x) \) is given by \( f'(x)g(x) + f(x)g'(x) \). Here:

• \(f'(x)\) is the derivative of \(f(x)\).
• \(g'(x)\) is the derivative of \(g(x)\).

Real-life calculus exercises often encounter these scenarios. This helps to break down complex expressions, like the one in our exercise, where applying the product rule leads to understanding and simplifying the components of derivatives. Mastering this ensures you can handle more intricate calculus challenges with ease.

Diving Deeper into Algebraic Expressions

Algebra is the cornerstone of mathematics, forming the basis for higher-level concepts, like those seen in calculus. When we talk about factoring an algebraic expression, it involves recognizing patterns and pulling out common factors. This can drastically simplify complex expressions and is an essential skill.

Consider the algebraic expression provided: \(3(2x-1)^2(2)(x+3)^{1/2} + (2x-1)^3\left(\frac{1}{2}\right)(x+3)^{-1/2}\)This expression may appear intimidating initially. However, by identifying common factors—like \((2x-1)^2(x+3)^{-1/2}\)—we can begin simplifying.

Breaking down terms helps to manage them more effectively, allowing us to combine like terms and see potential simplifications or factors. For instance, after finding that common factor, the expression becomes much easier to work with, leading to greater efficiency in solving problems. Recognizing these patterns comes with practice and builds the foundation for managing more advanced mathematical concepts seamlessly.

Calculus in Action: Realizing Applications

Calculus extends far beyond academic exercises and has practical, real-world applications in various fields. Engineers, scientists, and economists constantly rely on calculus to model and solve real-world problems.

The exercise we tackled shows a fundamental use of calculus beyond theoretical learning. Applications of this discipline are wide-ranging:

• In physics, to find the dynamics of moving objects by describing their motion.
• In economics, to optimize profit and minimize costs.
• In medicine, to model the growth of populations or the spread of diseases.

Understanding the calculus concepts applied in specific problems allows students to draw connections to fields they might explore in future careers. By seeing how calculus is used to factor complex expressions—like in our exercise—learners can anticipate its power and necessity in solving intricate challenges spanning various scientific and industrial domains.