Concept : Let `a_(0)x^(n)+a_(1)x^(n-1)+…+a_(n-1)x+a_(n)=0` be the nth degree equation with `a_(0),a_(1),…a_(n)` integers. If p/q is a rational root of this equation, then p is a divisor of `a_(n)` and q is a divisor of `a_(0)`. If `a_(0)=1`, then every rational root of this equation must be an integer.
The rational roots of the equation `3x^(3)-x^(2)-3x+1=0` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

To solve the equation \(3x^3 - x^2 - 3x + 1 = 0\) and find its rational roots, we will follow these steps:

### Step 1: Identify the coefficients
The coefficients of the polynomial \(3x^3 - x^2 - 3x + 1\) are:
- \(a_0 = 3\) (coefficient of \(x^3\))
- \(a_1 = -1\) (coefficient of \(x^2\))
- \(a_2 = -3\) (coefficient of \(x\))
- \(a_3 = 1\) (constant term)

### Step 2: Apply the Rational Root Theorem
According to the Rational Root Theorem, any rational root \(p/q\) of the polynomial must have \(p\) as a divisor of the constant term \(a_3\) and \(q\) as a divisor of the leading coefficient \(a_0\).

- Divisors of \(a_3 = 1\): \( \pm 1\)
- Divisors of \(a_0 = 3\): \( \pm 1, \pm 3\)

### Step 3: List possible rational roots
The possible rational roots (from the combinations of divisors of \(a_3\) and \(a_0\)) are:
- \(1\)
- \(-1\)
- \(\frac{1}{3}\)
- \(-\frac{1}{3}\)

### Step 4: Test each possible rational root
We will substitute each possible rational root into the polynomial to check if it equals zero.

1. **Testing \(x = 1\)**:
\[ 3(1)^3 - (1)^2 - 3(1) + 1 = 3 - 1 - 3 + 1 = 0 \quad \text{(Root found)} \]

2. **Testing \(x = -1\)**:
\[ 3(-1)^3 - (-1)^2 - 3(-1) + 1 = -3 - 1 + 3 + 1 = 0 \quad \text{(Root found)} \]

3. **Testing \(x = \frac{1}{3}\)**:
\[ 3\left(\frac{1}{3}\right)^3 - \left(\frac{1}{3}\right)^2 - 3\left(\frac{1}{3}\right) + 1 = 3\left(\frac{1}{27}\right) - \left(\frac{1}{9}\right) - 1 + 1 = \frac{1}{9} - \frac{1}{9} - 1 + 1 = 0 \quad \text{(Root found)} \]

4. **Testing \(x = -\frac{1}{3}\)**:
\[ 3\left(-\frac{1}{3}\right)^3 - \left(-\frac{1}{3}\right)^2 - 3\left(-\frac{1}{3}\right) + 1 = 3\left(-\frac{1}{27}\right) - \left(\frac{1}{9}\right) + 1 + 1 = -\frac{1}{9} - \frac{1}{9} + 1 + 1 \neq 0 \quad \text{(Not a root)} \]

### Step 5: Conclusion
The rational roots of the equation \(3x^3 - x^2 - 3x + 1 = 0\) are:
- \(x = 1\)
- \(x = -1\)
- \(x = \frac{1}{3}\)