(x+2)(x-3)(x-5) 0

4 min read Jun 16, 2024
(x+2)(x-3)(x-5) 0

Solving the Inequality: (x+2)(x-3)(x-5) > 0

This article will guide you through solving the inequality (x+2)(x-3)(x-5) > 0. We'll break down the steps and explain the reasoning behind each one.

Understanding the Problem

The inequality (x+2)(x-3)(x-5) > 0 asks us to find all the values of x that make the product of the three factors greater than zero.

Step 1: Finding the Critical Points

The critical points are the values of x that make the expression equal to zero. To find them, set each factor equal to zero and solve:

  • x + 2 = 0 => x = -2
  • x - 3 = 0 => x = 3
  • x - 5 = 0 => x = 5

These critical points divide the number line into four intervals:

  • Interval 1: x < -2
  • Interval 2: -2 < x < 3
  • Interval 3: 3 < x < 5
  • Interval 4: x > 5

Step 2: Testing the Intervals

Now, we'll pick a test value within each interval and substitute it into the original inequality to see if it satisfies the condition.

Interval 1 (x < -2): Let's choose x = -3

(-3 + 2)(-3 - 3)(-3 - 5) = (-1)(-6)(-8) = -48. This is negative, so the interval x < -2 does not satisfy the inequality.

Interval 2 (-2 < x < 3): Let's choose x = 0

(0 + 2)(0 - 3)(0 - 5) = (2)(-3)(-5) = 30. This is positive, so the interval -2 < x < 3 satisfies the inequality.

Interval 3 (3 < x < 5): Let's choose x = 4

(4 + 2)(4 - 3)(4 - 5) = (6)(1)(-1) = -6. This is negative, so the interval 3 < x < 5 does not satisfy the inequality.

Interval 4 (x > 5): Let's choose x = 6

(6 + 2)(6 - 3)(6 - 5) = (8)(3)(1) = 24. This is positive, so the interval x > 5 satisfies the inequality.

Step 3: Writing the Solution

We found that the intervals -2 < x < 3 and x > 5 satisfy the inequality (x+2)(x-3)(x-5) > 0.

Therefore, the solution to the inequality is:

-2 < x < 3 or x > 5

This can also be expressed in interval notation as:

(-2, 3) U (5, ∞)

Conclusion

By following these steps, we have successfully solved the inequality (x+2)(x-3)(x-5) > 0. We determined the critical points, tested the intervals, and expressed the solution both in inequality and interval notation.

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