Problem 25 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ (x-1)(x+2)(x-3)=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 1\), \(x = -2\), and \(x = 3\).

Step 1: Identify the Zero Product Property

The equation \((x-1)(x+2)(x-3)=0\) is a product of factors set equal to zero. According to the Zero Product Property, if the product of several factors is zero, at least one of the factors must be zero.

Step 2: Set Each Factor to Zero

Based on the Zero Product Property, set each factor equal to zero: \[ x-1 = 0, \]\[ x+2 = 0, \]\[ x-3 = 0. \]

Step 3: Solve Each Equation

Solve the equations from Step 2 separately:1. For \(x-1=0\): Add 1 to both sides to get \(x=1\).2. For \(x+2=0\): Subtract 2 from both sides to get \(x=-2\).3. For \(x-3=0\): Add 3 to both sides to get \(x=3\).

Step 4: Write the Solutions

The solutions to the equation \((x-1)(x+2)(x-3)=0\) are the values of \(x\) that make any of the factors zero. Thus, the solutions are \(x = 1\), \(x = -2\), and \(x = 3\).

Key Concepts

Factoring

Factoring is an important technique in algebra, especially when dealing with polynomial equations. It involves rewriting an expression as a product of its factors. For example, the equation \[(x-1)(x+2)(x-3)=0\]is already factored. This means we have broken down the complex expression into simpler, multiplicative components. Each of these is known as a factor.Understanding the factors is crucial. It helps us apply useful properties like the Zero Product Property. Factoring not only simplifies solving equations but also offers insight into the behavior of functions or expressions. When an expression is in factored form, recognizing the roots or solutions becomes straightforward.To apply factoring effectively:

• Identify common factors or patterns, such as trinomial squares.
• Break down expressions into the simplest components possible.
• Ensure that each term is double-checked for correct factoring.

Doing this correctly is your first step towards easier and more efficient problem-solving.

Solving Equations

Solving equations involves finding all possible values of a variable that make the equation true. When you have a factored equation like \[(x-1)(x+2)(x-3)=0\]applying the Zero Product Property is key. This property states that if a product of factors equals zero, at least one of the factors must be zero. Therefore, our task becomes breaking down the solution into simpler parts.For each factor, set it to zero:

• \(x-1=0\), solving gives \(x=1\).
• \(x+2=0\), solving gives \(x=-2\).
• \(x-3=0\), solving gives \(x=3\).

Each of these mini-equations is solved by simple algebraic manipulations.Remember, even as equations grow more complex, the principle remains the same. Break it down, solve individually, then combine your solutions. This systematic approach not only ensures accuracy but also builds strong foundational skills.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operators. They form the backbone of algebra and are used extensively in solving equations. In the equation\[(x-1)(x+2)(x-3)=0\]each part like \(x-1\) is a simple algebraic expression.Understanding these expressions:

• Variables: Symbols representing unknowns or quantities (like \(x\) in the equation).
• Operators: Symbols that perform operations (like addition or subtraction).
• Constants: Fixed values that do not change (like 1, 2, or 3 in the equation).

Manipulating these expressions effectively requires understanding their structure. Recognizing which operations to perform and in what order can simplify any algebra problem.Mastering algebraic expressions empowers you not only to solve equations but also to model real-world situations. The skills built here are fundamental for advanced mathematics and diverse applications in science and engineering.