We solve \( (x + 1)(x - 2)(x - 3) < 0 \) by analyzing the sign changes over intervals determined by the roots \( x = -1, 2, 3 \).
["# Solving the Inequality ( (x + 1)(x - 2)(x - 3) < 0 ) Using Sign Analysis", "Inequalities involving products of linear factors, such as ( (x + 1)(x - 2)(x - 3) < 0 ), are common in algebra and require careful analysis of the sign changes across critical points. In this article, we will solve the inequality step-by-step by identifying the roots and analyzing the sign of the expression in each interval.", "## Understanding the Inequality", "We aim to find all real values of ( x ) for which the product ( (x + 1)(x - 2)(x - 3) ) is negative (less than zero). The expression is a cubic polynomial, meaningful changes in sign occur at its roots, where each factor equals zero.", "The roots are:\n[\nx = -1, \quad x = 2, \quad x = 3\n]\nThese three values divide the real number line into four intervals:\n1. ( (-\infty, -1) )\n2. ( (-1, 2) )\n3. ( (2, 3) )\n4. ( (3, \infty) )", "We will determine the sign (positive or negative) of the product in each interval.", "## Step-by-Step Sign Analysis", "### 1. Interval ( (-\infty, -1) )", "Choose a test point less than (-1), e.g., ( x = -2 ):", "[\n(-2 + 1)(-2 - 2)(-2 - 3) = (-1)(-4)(-5) = -20\n]", "Result: Negative", "### 2. Interval ( (-1, 2) )", "Choose a test point between (-1) and (2), e.g., ( x = 0 ):", "[\n(0 + 1)(0 - 2)(0 - 3) = (1)(-2)(-3) = 6\n]", "Result: Positive", "### 3. Interval ( (2, 3) )", "Choose a test point between (2) and (3), e.g., ( x = 2.5 ):", "[\n(2.5 + 1)(2.5 - 2)(2.5 - 3) = (3.5)(0.5)(-0.5) = -0.875\n]", "Result: Negative", "### 4. Interval ( (3, \infty) )", "Choose a test point greater than (3), e.g., ( x = 4 ):", "[\n(4 + 1)(4 - 2)(4 - 3) = (5)(2)(1) = 10\n]", "Result: Positive", "## Summary of Signs", "| Interval | Sign of ( (x+1)(x-2)(x-3) ) |\n|---------------------|-----------------------------|\n| ( (-\infty, -1) ) | Negative |\n| ( (-1, 2) ) | Positive |\n| ( (2, 3) ) | Negative |\n| ( (3, \infty) ) | Positive |", "## Finding the Solution", "We seek where the expression is less than zero, i.e., negative. From the table, this occurs in:", "- ( (-\infty, -1) )\n- ( (2, 3) )", "Thus, the solution to ( (x + 1)(x - 2)(x - 3) < 0 ) is:", "[\nx \in (-\infty, -1) \cup (2, 3)\n]", "## Why This Method Works", "By analyzing sign changes across intervals defined by the roots, we exploit the continuous nature of polynomials. Since there are no repeated roots, each factor changes sign exactly once at its root. This sign analysis allows us to pinpoint intervals of positivity and negativity without solving the equation explicitly—just testing representative values confirms the trend.", "## Final Answer", "[\n\boxed{(x + 1)(x - 2)(x - 3) < 0 \quad \ ext{for} \quad x \in (-\infty, -1) \cup (2, 3)}\n]", "Understanding this sign behavior is essential for solving higher-degree polynomial inequalities efficiently. Combine this method with graphing or number line testing to master these concepts."]
