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Interval notation is a way to describe a set of numbers along a number line. It is used to express the solution to inequalities. In this approach, we use brackets and parentheses to show which numbers are included or excluded from the set.

• Square brackets, "[" or "]", mean that the endpoint is included in the interval.
• Round brackets, "(" or ")", mean the endpoint is not included.

Consider the inequality solution set \((-\infty, -\frac{2}{3}] \cup [\frac{3}{2}, \infty)\). Here, the interval \((-\infty, -\frac{2}{3}]\) means all numbers less than or equal to \(-\frac{2}{3}\), and the interval \([\frac{3}{2}, \infty)\) signifies all numbers greater than or equal to \(\frac{3}{2}\). The union symbol "\(\cup\)" connects disjoint intervals, indicating that any number falling in either interval makes the inequality true.

Factoring quadratics is a method used to simplify quadratic expressions, making it easier to solve equations or inequalities. The goal is to express a quadratic equation in the form \(ax^2 + bx + c\) as a product of two binomial expressions \((px + q)(rx + s)\).
For the expression \(6x^2 - 5x - 6\), find two numbers that multiply to the product of the leading coefficient \(6\) and constant term \(-6\), which is \(-36\), and sum to the middle coefficient \(-5\). These numbers are \(-9\) and \(4\).
Next, rewrite the equation to group terms: \(6x^2 - 9x + 4x - 6\), and proceed by factoring by grouping. It results in \((3x+2)(2x-3)\), highlighting how the original quadratic splits into multiplication of simpler factors.

Critical points in the context of quadratic inequalities are the values of \(x\) that make the expression equal zero. They divide the number line into intervals that can then be tested.
You find critical points by setting the factored expressions equal to zero and solving for \(x\).
For the inequality \((3x+2)(2x-3)\geq0\):

• Set \(3x+2=0\), solving gives \(x=-\frac{2}{3}\).
• Set \(2x-3=0\), solving gives \(x=\frac{3}{2}\).

These points \(-\frac{2}{3}\) and \(\frac{3}{2}\) are your critical points. They are essential in defining the intervals you need to test to determine where the inequality holds.

After determining the critical points, we must test the intervals between them to understand where the inequality is satisfied. Each interval is checked using a test point.

Start by identifying the intervals between the critical points: \(-\infty \to -\frac{2}{3}\), \(-\frac{2}{3}\) to \(\frac{3}{2}\), and \(\frac{3}{2} \to \infty\). For each interval, select a point to substitute back into the factored expression:

• For \((-\infty, -\frac{2}{3})\), test \(x = -1\): \((3(-1) + 2)(2(-1) - 3) = 5 > 0\), so this interval holds.
• For \((-\frac{2}{3}, \frac{3}{2})\), use \(x = 0\): \((3(0) + 2)(2(0) - 3) = -6 < 0\), showing this interval does not hold.
• For \((\frac{3}{2}, \infty)\), test \(x = 2\): \((3(2) + 2)(2(2) - 3) = 8 > 0\), confirming this interval holds.

This step efficiently determines which intervals satisfy the inequality, leading us to the solution in interval notation.