Problem 18 - Contemporary Precalculus

Question

Factor the given expression. For example, $$x-x^{1 / 2}-2=\left(x^{1 / 2}-2\right)\left(x^{1 / 2}+1\right)$$ $$x^{2 / 7}-2 x^{1 / 7}-15$$

Step-by-Step Solution

Answer

Question: Factor the expression $$x^{2/7}-2x^{1/7}-15$$. Answer: $$(x^{1/7} - 5)(x^{1/7} + 3)$$

Step 1: Perform substitution

Let $$u = x^{1/7}$$, then $$u^2 = x^{2/7}$$. The original expression becomes $$u^2 - 2u - 15$$

Step 2: Factor the new expression

Now, we need to factor the quadratic expression. $$(u - 5)(u + 3)$$ This is the factored form of the new expression in terms of "u".

Step 3: Substitute back and find the factored form of the original expression

We have $$u=x^{1/7}$$, substitute back to the factored expression: $$(x^{1/7} - 5)(x^{1/7} + 3)$$ This is the factored form of the original expression.

Key Concepts

Substitution Method

The substitution method is a powerful tool used in algebra to make expressions easier to handle. When faced with complex expressions, replacing parts of them with simpler symbols can simplify the problem. In this exercise, we tackled a seemingly complex expression by setting a substitution variable.

- We substituted \( u = x^{1/7} \) to transform \( x^{2/7} \) into \( u^2 \). This simplification is crucial because it converts a non-linear expression into a more familiar quadratic form.

- After substitution, the new expression is \( u^2 - 2u - 15 \), which is much easier to factor compared to the original expression.

The advantage of the substitution method is it allows us to temporarily work with simpler equations before going back to the original variables. It's important to always remember to substitute back the original terms to get the final factored expression.

Quadratic Expressions

Understanding quadratic expressions is fundamental in algebra. A quadratic expression is any expression that can be arranged in the form \( ax^2 + bx + c \). Quadratic expressions are common, and their characteristics make them easy to manipulate once you learn the basic techniques.

- In this exercise, we dealt with \( u^2 - 2u - 15 \), where the goal was to factor it into two binomials.

- Factoring involves finding two numbers that multiply to \( c \) (here, -15) and add to \( b \) (here, -2). The solution comes from identifying the numbers -5 and +3 because \( (-5) \times (+3) = -15 \) and \((-5) + (+3) = -2 \).

This results in the factors being \( (u-5)(u+3) \). Factoring quadratics is a repetitive skill that improves with practice and is integral in simplifying algebraic expressions in mathematics.

Mathematics Education

Learning math is not just about getting the correct answer; it's about understanding the concepts. Mathematics education emphasizes understanding the 'why' and 'how' behind each solution. This educational philosophy was demonstrated in the exercise through the clear use of the substitution method to simplify a complex algebraic expression.

- Through engaging with problems like these, students develop critical thinking skills. They learn to view complex problems from different angles and apply various methods to simplify their work.

- Exercises like this one highlight the connection between algebra and real-world problem-solving skills. By breaking down a problem into smaller, more manageable parts, students learn a crucial approach not only in mathematics but across different fields.

Ensuring that students understand each step and its purpose enhances their learning and helps build a stronger foundation for future, more advanced mathematical challenges.