Question

Алгебра (x+1)(x-2)(2x+5)>_0

Answer 1

Hello Friend.

x ∈ (-5\2 ; -1) ∪ (2; +∞)

Answer 2

To solve the inequality $(x+1)(x-2)(2x+5) \geq 0$, we need to find the intervals of values for $x$ that make the expression greater than or equal to zero.

To do this, we can use a sign chart or test points in the intervals. Let's go with the sign chart method.

Step 1: Find the critical points by setting each factor equal to zero and solving for $x$:

1. $x + 1 = 0 \Rightarrow x = -1$
2. $x - 2 = 0 \Rightarrow x = 2$
3. $2x + 5 = 0 \Rightarrow x = -\frac{5}{2}$

Step 2: Create a sign chart with the critical points and test points in each interval:

Interval 1: $(- \infty, -\frac{5}{2})$ Test point: $x = -3$

Interval 2: $(- \frac{5}{2}, -1)$ Test point: $x = -2$

Interval 3: $(-1, 2)$ Test point: $x = 0$

Interval 4: $(2, +\infty)$ Test point: $x = 3$

Step 3: Evaluate the expression $(x+1)(x-2)(2x+5)$ for each test point:

1. Test point $x = -3$: $(-3 + 1)(-3 - 2)(2(-3) + 5) = (-2)(-5)(-1) = 10 > 0$

2. Test point $x = -2$: $(-2 + 1)(-2 - 2)(2(-2) + 5) = (-1)(-4)(1) = 4 > 0$

3. Test point $x = 0$: $(0 + 1)(0 - 2)(2(0) + 5) = (1)(-2)(5) = -10 < 0$

4. Test point $x = 3$: $(3 + 1)(3 - 2)(2(3) + 5) = (4)(1)(11) = 44 > 0$

Step 4: Analyze the results:

The expression $(x+1)(x-2)(2x+5)$ is greater than or equal to zero in the intervals $(- \infty, -\frac{5}{2})$ and $(2, +\infty)$. So, the solution to the inequality is:

$x \in (-\infty, -\frac{5}{2}) \cup (2, +\infty)$

In interval notation, the solution is $(- \infty, -\frac{5}{2}) \cup (2, +\infty)$.