Sure, let's solve the inequality [tex]\((x+1)\left(x^2-3x+2\right)<0\)[/tex] step-by-step.
### Step 1: Factor the Quadratic Expression
First, factor the quadratic expression [tex]\(x^2 - 3x + 2\)[/tex].
[tex]\[
x^2 - 3x + 2 = (x-1)(x-2)
\][/tex]
So, the inequality becomes:
[tex]\[
(x + 1)(x - 1)(x - 2) < 0
\][/tex]
### Step 2: Determine Critical Points
Identify the critical points where each factor is zero:
1. [tex]\(x + 1 = 0\)[/tex] gives [tex]\(x = -1\)[/tex]
2. [tex]\(x - 1 = 0\)[/tex] gives [tex]\(x = 1\)[/tex]
3. [tex]\(x - 2 = 0\)[/tex] gives [tex]\(x = 2\)[/tex]
### Step 3: Test Intervals Between Critical Points
These critical points divide the real number line into four intervals:
- [tex]\((-\infty, -1)\)[/tex]
- [tex]\((-1, 1)\)[/tex]
- [tex]\((1, 2)\)[/tex]
- [tex]\((2, \infty)\)[/tex]
We need to test the sign of the expression [tex]\((x+1)(x-1)(x-2)\)[/tex] in each of these intervals.
1. Interval [tex]\((-\infty, -1)\)[/tex]:
Take [tex]\(x = -2\)[/tex]:
[tex]\[
(-2 + 1)(-2 - 1)(-2 - 2) = (-1)(-3)(-4) = -12 < 0
\][/tex]
This interval satisfies the inequality.
2. Interval [tex]\((-1, 1)\)[/tex]:
Take [tex]\(x = 0\)[/tex]:
[tex]\[
(0 + 1)(0 - 1)(0 - 2) = (1)(-1)(-2) = 2 > 0
\][/tex]
This interval does not satisfy the inequality.
3. Interval [tex]\((1, 2)\)[/tex]:
Take [tex]\(x = 1.5\)[/tex]:
[tex]\[
(1.5 + 1)(1.5 - 1)(1.5 - 2) = (2.5)(0.5)(-0.5) = -0.625 < 0
\][/tex]
This interval satisfies the inequality.
4. Interval [tex]\((2, \infty)\)[/tex]:
Take [tex]\(x = 3\)[/tex]:
[tex]\[
(3 + 1)(3 - 1)(3 - 2) = (4)(2)(1) = 8 > 0
\][/tex]
This interval does not satisfy the inequality.
### Step 4: Compile the Solution
The intervals that satisfy the inequality [tex]\((x + 1)(x - 1)(x - 2) < 0\)[/tex] are from [tex]\((-\infty, -1)\)[/tex] and [tex]\((1, 2)\)[/tex].
Thus, the solution is:
[tex]\[
x \in (-\infty, -1) \cup (1, 2)
\][/tex]
In terms of notation for your input format:
[tex]\[
-\infty < x < -1 \quad \text{or} \quad 1 < x < 2
\][/tex]
And there you have it!
