Problem 59 - Algebra A Combined Function

Question

Factor. $$ z^{3}-1 $$

Step-by-Step Solution

Answer

The factorization is \((z-1)(z^2+z+1)\).

Step 1: Identify the Expression Type

The expression given is a difference of two cubes. It can be rewritten as \( z^3 - 1^3 \). The difference of cubes formula is \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \). Here, \( a = z \) and \( b = 1 \).

Step 2: Apply the Difference of Cubes Formula

Substitute \( a = z \) and \( b = 1 \) into the difference of cubes formula. The expression becomes \( (z - 1)(z^2 + z \cdot 1 + 1^2) \).

Step 3: Simplify the Expression

Simplify the factored terms: \( (z - 1) \) remains the same, \( z^2 \) is as it is, \( z \cdot 1 = z \), and \( 1^2 = 1 \). So the expression simplifies to \( (z - 1)(z^2 + z + 1) \).

Step 4: Verify the Factorization

Expand \( (z - 1)(z^2 + z + 1) \) to verify the correctness. Multiply \( z \) and \(-1\) by each term in \( z^2 + z + 1 \) and simplify. It results back in \( z^3 - 1 \). This confirms that \( (z - 1)(z^2 + z + 1) \) is the correct factorization.

Key Concepts

Difference of Cubes

When you come across an expression like \( z^3 - 1 \), you're dealing with a special type of algebraic expression known as the "difference of cubes." This term functions like a key to unlock a simpler form through factorization.

A difference of cubes looks like \( a^3 - b^3 \) and can be rewritten as \( (a - b)(a^2 + ab + b^2) \). This formula helps us break down complex cubic terms into more manageable parts.

• The first term \( (a-b) \) denotes the simple subtraction of cube roots.
• The second component \( (a^2 + ab + b^2) \) accounts for the remaining parts of the cubic expression.

In this specific exercise, we identify \( a \) as \( z \) and \( b \) as \( 1 \). By substituting these variables into the formula, our expression becomes \( (z - 1)(z^2 + z + 1) \).

It's important to practice recognizing the difference of cubes structure as it frequently appears in algebra, helping simplify polynomials more efficiently.

Algebraic Expressions

Algebraic expressions like \( z^3 - 1 \) are mathematical phrases that can consist of variables and constants. They're combined using operations and represent specific values or rules.

Understanding how to manipulate these expressions is crucial because they're foundational in mathematics, allowing us to perform calculations, solve equations, and model real-world scenarios.

• Variables represent unknown values and are often denoted by letters such as \( x, y, z \), etc.
• Constants are specific, fixed numbers in the expression such as \( 1, 2, 3, \ldots \) and so on.
• Operations, such as addition, subtraction, multiplication, and division, help combine variables and constants.

In the case of \( z^3 - 1 \), you see the variable \( z \) raised to a power and a constant \( 1 \). Recognizing the pattern within the expression helps determine the appropriate factorization method. Breaking down these expressions step-by-step, using formulas like the difference of cubes, simplifies them into factors. This process is key in algebra, making further calculations much easier.

Polynomials

Polynomials are a type of algebraic expression that you'll encounter frequently. They consist of terms combined by addition or subtraction, where each term is a product of constants and variables raised to whole number powers.

The expression \( z^3 - 1 \) is a polynomial, specifically a cubic polynomial since the highest power is three. These types of polynomials are important when studying functions and solving equations.

• Cubic polynomials, like \( z^3 - 1 \), involve a variable raised to the third power.
• They can be factored using special formulas like the difference of cubes.
• Factoring simplifies these expressions, revealing solutions or roots of polynomial equations.

Recognizing polynomials and knowing how to manipulate them is crucial for solving algebraic equations. In this problem, simplifying \( z^3 - 1 \) using the difference of cubes gives us \( (z - 1)(z^2 + z + 1) \). Factored forms provide simpler representations of equations, assisting us in solving for unknowns efficiently. Understanding how to work with polynomials enhances your overall algebraic proficiency, building the foundation for more advanced mathematical concepts.