(3x+10)(2x^2+1)(2-x) 0

3 min read Jun 16, 2024
(3x+10)(2x^2+1)(2-x) 0

Solving the Inequality: (3x + 10)(2x² + 1)(2 - x) > 0

This inequality involves a product of three factors, and we need to determine the intervals where the product is positive. Here's a step-by-step approach:

1. Find the Critical Points

The critical points are the values of x that make one or more of the factors equal to zero.

  • 3x + 10 = 0 => x = -10/3
  • 2x² + 1 = 0 => This equation has no real solutions because 2x² is always non-negative and adding 1 makes it always positive.
  • 2 - x = 0 => x = 2

Therefore, our critical points are x = -10/3 and x = 2.

2. Create a Sign Chart

The critical points divide the number line into three intervals:

  • x < -10/3
  • -10/3 < x < 2
  • x > 2

We'll create a sign chart to determine the sign of each factor in each interval:

Interval 3x + 10 2x² + 1 2 - x (3x + 10)(2x² + 1)(2 - x)
x < -10/3 - + + -
-10/3 < x < 2 + + + +
x > 2 + + - -

Explanation:

  • 3x + 10: This factor is negative when x < -10/3 and positive when x > -10/3.
  • 2x² + 1: This factor is always positive.
  • 2 - x: This factor is positive when x < 2 and negative when x > 2.

3. Determine the Solution

We want to find the intervals where the product of the factors is positive. Looking at the sign chart, the product is positive in the interval:

-10/3 < x < 2

4. Final Answer

The solution to the inequality (3x + 10)(2x² + 1)(2 - x) > 0 is -10/3 < x < 2.

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