how do you solve:

x²-2x-15 ≤ 0

please help :)

oh so that's it? lol i did that but i thought it wasn't correct and i've been trying other methods for the past hour. thatnx alot :)

LOL..you're welcome

To solve the inequality x²-2x-15 ≤ 0, you can use a method called factoring.

Step 1: Set the inequality to equal to 0.

x²-2x-15 = 0

Step 2: Factor the quadratic equation.

The equation can be factored as (x-5)(x+3) = 0

Step 3: Find the critical points.

The critical points are the values of x where the equation equals 0. In this case, the critical points are x = 5 and x = -3.

Step 4: Plot the critical points on a number line.

On a number line, mark the critical points x = 5 and x = -3.

---o---o---

Step 5: Test a value from each interval to determine the solution.

Now, select test points from each interval that are not critical points. For simplicity, you can choose numbers like -4, 0, and 6.

For interval (-∞, -3), we can choose x = -4. Substituting this value in the original inequality: (-4-5)(-4+3) ≤ 0, we get (-9)(-1) ≤ 0, which is true.

For interval (-3, 5), we can choose x = 0. Substituting this value in the original inequality: (0-5)(0+3) ≤ 0, we get (-5)(3) ≤ 0, which is false.

For interval (5, +∞), we can choose x = 6. Substituting this value in the original inequality: (6-5)(6+3) ≤ 0, we get (1)(9) ≤ 0, which is false.

Step 6: Determine the solution.

Any interval where the inequality is true will fall into the solution set. Therefore, (-∞, -3] is included in the solution set.

Thus, the solution to the inequality x²-2x-15 ≤ 0 is x ≤ -3.

You want to get two numbers that multiply to the last term (-15) and add to the middle term (-2). These numbers are -5 and 3. (x-5)(x+3) is your answer