Problem 22 - Contemporary Precalculus

Question

Factor the given expression. For example, $$x-x^{1 / 2}-2=\left(x^{1 / 2}-2\right)\left(x^{1 / 2}+1\right)$$ $$x+3 x^{2 / 3}+3 x^{1 / 3}+1$$

Step-by-Step Solution

Answer

Answer: The factored form of the expression is $$(x^{\frac{1}{3}}+1)^3$$.

Step 1: Observe the pattern of exponents in the expression

Notice that the exponents in the expression are $$\frac{2}{3}$$ and $$\frac{1}{3}$$. These exponents share a common denominator (3), which suggests that we can make a substitution to make the expression easier to work with.

Step 2: Make a substitution

Let $$y = x^{\frac{1}{3}}$$. Then, $$y^2 = x^{\frac{2}{3}}$$. Our expression now becomes: $$x + 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} + 1 = y^3 + 3y^2 + 3y + 1$$

Step 3: Factor the expression with the substitution

We can recognize that this expression is a perfect cube trinomial. The general form for a perfect cube trinomial is $$(a+b)^3 = a^3 + 3a^2b+3ab^2+b^3$$ Comparing this with our expression, we can see that $$a=y$$ and $$b=1$$. Thus, we can rewrite the expression as $$(y+1)^3$$.

Step 4: Substitute back the original variable

Replace $$y$$ with $$x^{\frac{1}{3}}$$, so the factored expression is: $$(x^{\frac{1}{3}}+1)^3$$ So the factored form of the original expression is: $$x+3 x^{\frac{2}{3}}+3 x^{\frac{1}{3}}+1 = \left(x^{\frac{1}{3}}+1\right)^3$$

Key Concepts

Precalculus

Precalculus bridges the gap between algebra and calculus, preparing you for more advanced mathematical concepts. In precalculus, you'll deal with complex expressions and learn techniques for simplifying and solving them. Key areas of study in this subject include:

• Functions and their properties
• Polynomials and rational expressions
• Exponential and logarithmic expressions
• Trigonometric identities and equations

Understanding precalculus is essential because it lays the groundwork for calculus concepts such as limits, derivatives, and integrals. When working with problems that involve factoring expressions, it's crucial to identify patterns and techniques that simplify the process. By mastering precalculus, you're setting yourself up for success in calculus and other mathematical fields.

Exponents

Exponents are a fundamental concept in mathematics, representing repeated multiplication of a number by itself. For example, \( x^3 \) means \( x \) multiplied by itself three times: \( x \cdot x \cdot x \). In the context of the original problem, we encounter fractional exponents:

• \( x^{\frac{2}{3}} \)
• \( x^{\frac{1}{3}} \)

Fractional exponents indicate roots as well as powers. The expression \( x^{\frac{1}{3}} \) represents the cube root of \( x \). From the exercise, recognizing the pattern in exponents allows us to use substitutions, simplifying complex expressions into more manageable forms. This is a pivotal step when factoring the given problem.

Substitution Method

The substitution method is a powerful technique used to simplify problem-solving, especially in equations and expressions involving exponents. By temporarily replacing a complex part of the expression with a simpler variable, we can solve or transform equations more easily.
In our example, we substitute \( y = x^{\frac{1}{3}} \). Consequently, \( y^2 = x^{\frac{2}{3}} \). This transforms our expression into:

• \( y^3 + 3y^2 + 3y + 1 \)

After working through the simpler problem, we reverse the substitution to return to the original variables. This technique significantly eases the process of identifying and applying patterns such as perfect cube trinomials.

Perfect Cube Trinomial

A perfect cube trinomial is a type of expression based on the formula \[ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]
This identity is handy in factoring expressions that fit this particular pattern. In our example, the transformed expression \( y^3 + 3y^2 + 3y + 1 \) resembles the perfect cube trinomial form, where \( a = y \) and \( b = 1 \). This allows us to write it as \( (y+1)^3 \).
Understanding perfect cube trinomials helps in recognizing and factoring polynomials efficiently, especially those that might initially appear intimidating. Once you identify the pattern, the simplification becomes straightforward, leading to the correct factored form.