Problem 94 - Algebra and Trigonometry

Question

Factor the expression completely. (This type of expression arises in calculus in using the “product rule.”) $$ \frac{1}{3}(x+6)^{-2 / 3}(2 x-3)^{2}+(x+6)^{1 / 3}(2)(2 x-3)(2) $$

Step-by-Step Solution

Answer

Factor: \((x+6)^{-2/3}(2x-3)\left( \frac{2x}{3} - 1 + 4(x+6)^{1/3} \right)\).

Step 1: Identify Common Factors

First, observe that both terms have a common factor of \((x+6)^{-2/3}\) and \((2x-3)\). This simplifies the factorization process since we can factor these out from both terms.

Step 2: Factor Out the Common Factor

Factor \((x+6)^{-2/3}(2x-3)\) from both terms in the expression. This gives:\[\frac{1}{3}(x+6)^{-2/3}(2x-3)^{2} + 2(x+6)^{1/3}(2)(2x-3) = (x+6)^{-2/3}(2x-3)\left( \frac{1}{3}(2x-3) + 4(x+6)^{1/3} \right)\]

Step 3: Simplify the Expression Inside Parentheses

Simplify further inside the parentheses:\[\frac{1}{3}(2x-3) + 4(x+6)^{1/3}\]Multiplying, we have:\[\frac{1}{3}(2x-3) = \frac{2x}{3} - 1\]Thus:\[(x+6)^{-2/3}(2x-3)\left( \frac{2x}{3} - 1 + 4(x+6)^{1/3} \right)\]

Step 4: Verify and Rearrange if Necessary

Re-check the factored and simplified form to ensure no further simplifications are possible. The factorization is complete when all common factors are factored out. After verifying, you'll have:\[(x+6)^{-2/3}(2x-3)\left( \frac{2x}{3} - 1 + 4(x+6)^{1/3} \right)\] as the completely factored expression.

Key Concepts

Common Factors

Understanding common factors is essential when factoring expressions, especially those encountered in calculus. A common factor is a term that is shared by all parts of an expression. By identifying and factoring out these common factors, you simplify the entire expression, making it easier to work with and understand.

In the provided exercise, observe that both terms include

• each term has \((x+6)^{-2/3}\) and \((2x-3)\) as a factor.
• This makes it straightforward to extract these from the expression.

By pulling out the common factors, you reduce complexity and create a simpler expression that highlights the remaining terms needing further simplification. This is a fundamental technique, not just in basic algebra but also in higher-level mathematics like calculus. Recognizing and factoring out common factors is crucial in making intricate problems more manageable.

Expression Simplification

Once common factors are extracted, simplifying the expression further helps achieve a cleaner form. Simplification involves reducing the complexity of mathematical expressions.

In the example provided, after the common factor is factored out, the expression inside the parentheses needs to be simplified. This requires:

• Distributing constants across terms.
• Managing coefficients to present the expression in a reduced format.

One approach is to interpret and rearrange terms systematically, like converting \(\frac{1}{3}(2x-3)\) into \(\frac{2x}{3} - 1\). It shows how fractional coefficients can be rewritten for clarity.

Simplification is often about breaking down terms and recombining them in an insightful manner. This can involve combining like terms, clearing out fractions, and restructuring the components to maintain equality while making it look simpler and more elegant.

Calculus Product Rule

The calculus product rule is a way to find the derivative of a product of two functions. Importantly, this is where factorization techniques become valuable. In calculus, expressions like those given in the exercise frequently arise when applying the product rule.

The product rule states that if you have two functions, \(u(x)\) and \(v(x)\), their derivative is given by:
\[ (uv)' = u'v + uv' \]

In the context of the given expression, the use of common factors connects directly to how terms might appear when differentiating a product of functions. Understanding how to factor and simplify expressions can significantly ease dealing with product rule derivatives.

• This reduces efforts in handling terms and anticipating results correctly.
• Having a firm grasp of factorization and simplification supports comprehension and application of calculus concepts such as the product rule.

This exercise offers a practical insight into the intersection of algebraic manipulation and calculus operations, emphasizing the importance of mastering previous mathematical concepts for understanding advanced calculus topics.