Question

(х+3)(х-2)(х+1) > равно 0

Answer 1

Ответ прикреплен))))

Answer 2

(x+3)(x-2)(x+1)>0
найдем корни уравнения
(x+3)(x-2)(x+1)=0
x1=-3 x2=-1 x3=2
дальше все это наносим на числовую ось
и в итоге ответ: -3<x<-1; x>2 или x ∈ (-3;-1) ∪ (2; +∞)

Answer 3

To find the solution to the inequality $(x+3)(x-2)(x+1) > 0$, we need to determine the intervals where the expression is greater than zero.

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

1. $x + 3 = 0 \Rightarrow x = -3$
2. $x - 2 = 0 \Rightarrow x = 2$
3. $x + 1 = 0 \Rightarrow x = -1$

Step 2: Create a number line and mark these critical points:

pythonCopy code
...   -3   -2   -1    0    1    2   3   ...

Step 3: Test each interval:

1. Interval $(- \infty, -3)$: Pick $x = -4$. Check: $(-1)(-6)(-3) = -18 < 0$. This interval is not part of the solution.
2. Interval $(-3, -1)$: Pick $x = -2$. Check: $(1)(0)(-1) = 0 < 0$. This interval is not part of the solution.
3. Interval $(-1, 2)$: Pick $x = 0$. Check: $(3)(-2)(1) = -6 < 0$. This interval is not part of the solution.
4. Interval $(2, \infty)$: Pick $x = 3$. Check: $(6)(1)(4) = 24 > 0$. This interval is part of the solution.

Step 4: Combine the intervals:

The solution to the inequality $(x+3)(x-2)(x+1) > 0$ is $x > 2$ or, in interval notation, $(2, \infty)$.