Question

(x + 2) * x * (x - 1) * (x + 2) < 0

Answer 1

(x+2)*x*(x-1)*(x+2)<0
x=-2     x=0      x=1
   +                   +               -                  +
------- -2 ----------- 0 ---------- 1 ---------------
                                 \\\\\\\\\\\\

x= 2    +    +    +    +  | +
x=0.5  +    +    -     +  | -
x= -1   +    -    -      +  | +
x= -3   -    -     -      -  | +

x∈(0; 1)

Answer 2

To solve the inequality $(x + 2) \cdot x \cdot (x - 1) \cdot (x + 2) < 0$, you can use a sign chart or interval notation.

First, let's find the critical points where the expression equals zero:

1. $(x + 2) = 0$ when $x = -2$.
2. $x = 0$
3. $(x - 1) = 0$ when $x = 1$.

These critical points divide the real number line into four intervals:

Interval 1: $(-∞, -2)$ Interval 2: $(-2, 0)$ Interval 3: $(0, 1)$ Interval 4: $(1, ∞)$

Now, we'll test a point from each interval in the inequality to determine the sign of the expression in each interval:

1. For Interval 1, pick $x = -3$, which is in $(-∞, -2)$: $((-3 + 2) \cdot (-3) \cdot (-3 - 1) \cdot (-3 + 2)) < 0$ simplifies to $(-12) < 0$, which is true. So, Interval 1 is where the expression is negative.

2. For Interval 2, pick $x = -1$, which is in $(-2, 0)$: $((-1 + 2) \cdot (-1) \cdot (-1 - 1) \cdot (-1 + 2)) < 0$ simplifies to $(2) < 0$, which is false. So, Interval 2 is where the expression is positive.

3. For Interval 3, pick $x = 0.5$, which is in $(0, 1)$: $((0.5 + 2) \cdot (0.5) \cdot (0.5 - 1) \cdot (0.5 + 2)) < 0$ simplifies to $(3.75) < 0$, which is false. So, Interval 3 is where the expression is positive.

4. For Interval 4, pick $x = 2$, which is in $(1, ∞)$: $((2 + 2) \cdot (2) \cdot (2 - 1) \cdot (2 + 2)) < 0$ simplifies to $(32) < 0$, which is false. So, Interval 4 is where the expression is positive.

Now, summarize the results:

• The expression is negative on Interval 1: $(-∞, -2)$.
• The expression is positive on Intervals 2, 3, and 4: $(-2, 0)$, $(0, 1)$, and $(1, ∞)$.

So, the solution to the inequality $(x + 2) \cdot x \cdot (x - 1) \cdot (x + 2) < 0$ is:

$(-∞, -2) \cup (0, 1)$