Answer: x ∈ {-0.5, -5, -12.5}
Step-by-step explanation: To solve the inequality 2x³ + 21x² + 60x + 25 > 0, we first need to find the values of x that make the inequality true. We can do this by setting the expression equal to 0 and solving for x.
We can start by factoring the expression to make it easier to solve. Notice that 2x³ + 21x² + 60x + 25 is a polynomial with a leading coefficient of 2 and a constant term of 25. This means that it has the form (x + a)(x + b)(x + c), where a, b, and c are constants.
We can start by factoring out the common factor of 2x from the first two terms: 2x³ + 21x² + 60x + 25 = 2x(x² + 10.5x + 12.5). Now we can see that the expression has the form (x + a)(x + b)(x + c), where a = 0.5, b = 5, and c = 12.5.
So, we can rewrite the expression as (x + 0.5)(x + 5)(x + 12.5) = 0. Now we can solve for x by setting each factor equal to 0 and solving for x:
x + 0.5 = 0 => x = -0.5
x + 5 = 0 => x = -5
x + 12.5 = 0 => x = -12.5
Therefore, the values of x that make the inequality true are x = -0.5, x = -5, and x = -12.5.
Now we need to determine which of these values make the inequality 2x³ + 21x² + 60x + 25 > 0 true. To do this, we can substitute each of the values of x into the inequality and see which ones make the inequality true.
When x = -0.5, the inequality becomes 2(-0.5)³ + 21(-0.5)² + 60(-0.5) + 25 > 0, which simplifies to -0.5 + 5.25 - 15 + 25 > 0. This is true, because the left-hand side is 29 > 0.
When x = -5, the inequality becomes 2(-5)³ + 21(-5)² + 60(-5) + 25 > 0, which simplifies to -125 + 525 - 300 + 25 > 0. This is also true, because the left-hand side is 225 > 0.
When x = -12.5, the inequality becomes 2(-12.5)³ + 21(-12.5)² + 60(-12.5) + 25 > 0, which simplifies to -391.25 + 1181.25 - 750 + 25 > 0. This is also true, because the left-hand side is 1147.5 > 0.
Therefore, the solution to the inequality is x ∈ {-0.5, -5, -12.5}. This means that the values of x that make the inequality true are x = -0.5, x = -5, and x = -12.5. The inequality is satisfied when x is any of these values.