91Ó°ÊÓ

Factoring a quadratic equation is the first step in solving many polynomial inequalities. Let's take the example given in the exercise: \(x^2 - x - 2 > 0\). The goal here is to find numbers that multiply to give the constant term (-2) and add up to the linear coefficient (-1). In this case, those numbers are -2 and 1, since (-2) * (1) = -2 and (-2) + 1 = -1. So, we can factor the quadratic as \((x-2)(x+1)\). This factored form helps simplify the problem, making it easier to solve.

After factoring the quadratic expression, the next step involves finding the critical points. Critical points are the values of \(x\) that make each factor equal to zero. For the factored equation \((x-2)(x+1)\), we set each factor to zero: \(x-2=0\) and \(x+1=0\). Solving these gives \(x=2\) and \(x=-1\). These critical points divide the number line into separate intervals and play a crucial role in testing where the inequality holds true.

Once you've identified the critical points, you need to determine in which intervals the inequality holds. In this case, the critical points \(-1\) and \(2\) divide the number line into three intervals: \((-\infty, -1)\), \((-1, 2)\), and \((2, \infty)\). To test these intervals, pick a sample point from each one:
- For the interval \((-\infty, -1)\), choose \(x = -2\). Substituting \(x = -2\) into \((x-2)(x+1)\) results in \(4 > 0\).
- For the interval \((-1, 2)\), choose \(x = 0\). Substituting \(x = 0\) into \((x-2)(x+1)\) results in \(-2 < 0\).
- For the interval \((2, \infty)\), choose \(x = 3\). Substituting \(x = 3\) into \((x-2)(x+1)\) results in \(4 > 0\).

Polynomial inequalities are more complex than linear ones because they can cross the x-axis multiple times. The basic strategy for solving them is similar to what's shown in the exercise. You
- Factor the polynomial.
- Find the critical points by setting each factor to zero.
- Use these points to divide the number line into intervals.
- Test each interval by choosing a sample point and checking if the inequality is satisfied.
Combining the intervals that satisfy the inequality gives you the solution set. In our sample exercise, the solution is \(x \in (-\infty, -1) \cup (2, \infty)\). This approach can be applied universally to solve polynomial inequalities, making it a valuable tool in algebra.