Problem 41 - Algebra 2

Question

Solve each inequality using a graph, a table, or algebraically. $$ (x-1)(x+4)(x-3)>0 $$

Step-by-Step Solution

Answer

The solution is \(x \in (-4, 1) \cup (3, \infty)\).

Step 1: Find the critical points

To solve the inequality \((x-1)(x+4)(x-3) > 0\), we first find the critical points where the expression equals zero. Set each factor equal to zero: \(x-1=0\), \(x+4=0\), \(x-3=0\). Solving these equations, we get the critical points as \(x=1\), \(x=-4\), and \(x=3\).

Step 2: Determine the intervals

The critical points divide the number line into four intervals: \((-\infty, -4)\), \((-4, 1)\), \((1, 3)\), and \((3, \infty)\). We will test each interval to determine where the inequality holds.

Step 3: Test each interval

Choose test points from each interval and substitute them into \((x-1)(x+4)(x-3)\) to check the sign of the product: - Interval \((-\infty, -4)\): Use \(x=-5\). Substitute to get \((-5-1)(-5+4)(-5-3) = (-6)(-1)(-8) = -48\). Negative.- Interval \((-4, 1)\): Use \(x=0\). Substitute to get \((0-1)(0+4)(0-3) = (-1)(4)(-3) = 12\). Positive.- Interval \((1, 3)\): Use \(x=2\). Substitute to get \((2-1)(2+4)(2-3) = (1)(6)(-1) = -6\). Negative.- Interval \((3, \infty)\): Use \(x=4\). Substitute to get \((4-1)(4+4)(4-3) = (3)(8)(1) = 24\). Positive.

Step 4: Determine solution intervals

From the testing in Step 3, the solution occurs where the product is positive. This happens in the intervals \((-4, 1)\) and \((3, \infty)\).

Step 5: Write the solution

Combine the intervals where the inequality holds to write the final solution: \((x-1)(x+4)(x-3) > 0\) for \(x \in (-4, 1) \cup (3, \infty)\).

Key Concepts

Critical Points in Inequalities

When solving inequalities like \((x-1)(x+4)(x-3) > 0\), one of the most important steps is finding the critical points. Critical points are the values of \(x\) where the expression changes sign. They are found by setting each factor of the inequality equal to zero.

For this inequality, you have three factors: \(x-1\), \(x+4\), and \(x-3\). Set each of these equal to zero:

• \(x-1=0\) gives \(x=1\)
• \(x+4=0\) gives \(x=-4\)
• \(x-3=0\) gives \(x=3\)

Once found, these critical points \(-4\), \(1\), and \(3\) are placed on the number line. These points help to divide the number line into separate intervals for testing purposes.

Interval Testing: Deciding the Sign

Interval testing involves choosing test points from the subintervals formed by the critical points to determine where the inequality is true.

The inequalities can be broken down into intervals:

• \((-\infty, -4)\)
• \((-4, 1)\)
• \((1, 3)\)
• \((3, \infty)\)

Test points are selected from each interval and substituted back into the inequality \((x-1)(x+4)(x-3)\). This substitution helps to check whether the sign of the product is positive or negative.

For instance:

• From \((-\infty, -4)\), choose \(x = -5\).Substitute into the expression to find it yields a negative result.
• From \((-4, 1)\), choose \(x = 0\). Substituting in gives a positive result, indicating the inequality holds true in this interval.
• Continue this method for the other intervals, checking the resulting signs.

Solution Intervals for Inequalities

The solution intervals are the regions in which the tested values satisfy the inequality.

Using the result of interval testing from previous steps, you can combine the intervals where the product was positive to form the solution set of the inequality.

For our inequality \((x-1)(x+4)(x-3) > 0\), the tested intervals were:

• Positive in \((-4, 1)\)
• Negative in \((1, 3)\)
• Positive again in \((3, \infty)\)

The final solution would be the union of the positive intervals \(x \in (-4, 1) \cup (3, \infty)\). These are the ranges where the inequality holds true.