To solve the inequality \((2x² + 3x + 1)^x < 1\), let's analyze the behavior of the function and the inequality step by step.

### Step 1: Analyze the expression \( (2x^2 + 3x + 1) \)
We begin by examining the quadratic expression \( 2x^2 + 3x + 1 \). To do this, let's find its roots using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 2 \), \( b = 3 \), and \( c = 1 \). Plugging in these values:

\[
x = \frac{-3 \pm \sqrt{3^2 - 4(2)(1)}}{2(2)} = \frac{-3 \pm \sqrt{9 - 8}}{4} = \frac{-3 \pm \sqrt{1}}{4}
\]

\[
x = \frac{-3 \pm 1}{4}
\]

So, the roots are:

\[
x = \frac{-2}{4} = -\frac{1}{2} \quad \text{and} \quad x = \frac{-4}{4} = -1
\]

The roots of the quadratic expression \(2x^2 + 3x + 1\) are \(x = -\frac{1}{2}\) and \(x = -1\). This means that the quadratic expression factors as:

\[
2x^2 + 3x + 1 = 2(x + 1)(x + \frac{1}{2})
\]

### Step 2: Behavior of the quadratic expression
The quadratic expression \(2(x + 1)(x + \frac{1}{2})\) is positive when \(x\) is less than \(-1\) or greater than \(-\frac{1}{2}\), and it is negative between the roots \(-1 < x < -\frac{1}{2}\).

### Step 3: Consider the entire inequality
For the inequality \((2x^2 + 3x + 1)^x < 1\):

- **Case 1: \(x < 0\):** For negative values of \(x\), \(a^x\) is less than 1 if \(0 < a < 1\) or if \(a > 1\). Since \(2x^2 + 3x + 1\) is positive outside the interval \([-1, -\frac{1}{2}]\), but in this case, \(a^x\) where \(a > 1\) can be less than 1.
- **Case 2: \(x = 0\):** The expression \(a^x\) equals 1, and hence equality holds here.
- **Case 3: \(x > 0\):** For positive values of \(x\), \(a^x < 1\) implies \(0 < a < 1\). So, the expression \(2x^2 + 3x + 1\) must be between 0 and 1.

### Step 4: Find intervals
Considering all cases, we conclude:

1. **If \(x > 0\):** The inequality \( (2x^2 + 3x + 1)^x < 1 \) is true when \(0 < 2x^2 + 3x + 1 < 1\).
2. **If \(x < 0\):** The inequality holds when \(2x^2 + 3x + 1 > 1\).

Thus, the solution to the inequality \((2x² + 3x + 1)^x < 1\) depends on finding specific intervals of \(x\) based on these conditions. The intervals can be identified and tested through additional algebraic manipulation and interval testing.

Would you like the detailed calculation for these intervals?

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### Related Questions:
1. How do we determine the exact intervals where \(0 < 2x^2 + 3x + 1 < 1\)?
2. How do we factor a quadratic expression more efficiently?
3. What happens if \(x\) is very large or very small? What does that tell us about the inequality?
4. Can you show graphically where \( (2x^2 + 3x + 1)^x = 1\)?
5. How does the sign of \(x\) affect the solution set?

### Tip:
Remember, understanding the sign of a function and how exponents affect inequality is crucial in solving inequalities involving expressions raised to a variable exponent.

Learn how to solve the inequality (2x² + 3x + 1)^x < 1 using algebraic techniques and analysis of exponential functions. This problem covers quadratic equations, inequalities, and exponential properties, ideal for students in Grades 10-12.

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