The equation `x^(3) - 3x + 1 = 0` has

Options

A. no rational but three irrational roots
B. one rational and two irrational roots
C. no real roots
D. three rational roots

Explanation

To determine the nature of the roots of the equation \( x^3 - 3x + 1 = 0 \), we will follow these steps:

### Step 1: Check for Rational Roots
We will use the Rational Root Theorem, which states that any rational solution \( \frac{p}{q} \) of the polynomial equation \( ax^n + bx^{n-1} + \ldots + k = 0 \) must have \( p \) as a factor of the constant term \( k \) and \( q \) as a factor of the leading coefficient \( a \).

In our case, the polynomial is \( x^3 - 3x + 1 \):
- The constant term \( k = 1 \) (factors: \( \pm 1 \))
- The leading coefficient \( a = 1 \) (factors: \( \pm 1 \))

Thus, the possible rational roots are \( \pm 1 \).

### Step 2: Test Possible Rational Roots
We will substitute \( x = 1 \) and \( x = -1 \) into the equation to check if they are roots.

1. **For \( x = 1 \)**:
\[ f(1) = 1^3 - 3(1) + 1 = 1 - 3 + 1 = -1 \quad (\text{not a root}) \]

2. **For \( x = -1 \)**:
\[ f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3 \quad (\text{not a root}) \]

Since neither \( 1 \) nor \( -1 \) satisfies the equation, we conclude that there are no rational roots.

### Step 3: Determine the Nature of the Roots
Since the polynomial is a cubic equation, it must have three roots (real or complex). We have established that there are no rational roots.

To further analyze the nature of the roots, we can examine the derivative of the polynomial to find critical points and evaluate the behavior of the function.

### Step 4: Find the Derivative
The derivative of \( f(x) = x^3 - 3x + 1 \) is:
\[ f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1) \]

### Step 5: Find Critical Points
Setting the derivative to zero:
\[ 3(x - 1)(x + 1) = 0 \implies x = 1 \text{ or } x = -1 \]

### Step 6: Evaluate the Function at Critical Points
Now, we evaluate the function at the critical points:
1. **At \( x = 1 \)**:
\[ f(1) = -1 \]

2. **At \( x = -1 \)**:
\[ f(-1) = 3 \]

### Step 7: Analyze the Behavior of the Function
- As \( x \to -\infty \), \( f(x) \to -\infty \).
- At \( x = -1 \), \( f(-1) = 3 \).
- At \( x = 1 \), \( f(1) = -1 \).
- As \( x \to \infty \), \( f(x) \to \infty \).

Since the function changes signs between \( x = -1 \) and \( x = 1 \), and since it is a cubic polynomial, we can conclude that there are three real roots.

### Conclusion
The equation \( x^3 - 3x + 1 = 0 \) has **three real roots**, and since we have ruled out rational roots, these must be **irrational roots**.

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