Let `a_0x^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 arational root of this equation, then p is a divisor of an and q is a divisor of `a_n`. If `a_0 = 1`, then every rationalroot of this equation must be an integer.

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To solve the given problem, we need to analyze the cubic equation:

\[ x^3 - 13x^2 + 15x + 189 = 0 \]

### Step 1: Identify the equation
The equation is already given as:

\[ x^3 - 13x^2 + 15x + 189 = 0 \]

### Step 2: Find the factors of the constant term
The constant term \( a_n \) in the equation is 189. We need to find the integer factors of 189. The factors of 189 are:

- \( \pm 1 \)
- \( \pm 3 \)
- \( \pm 7 \)
- \( \pm 9 \)
- \( \pm 21 \)
- \( \pm 27 \)
- \( \pm 63 \)
- \( \pm 189 \)

### Step 3: Apply the Rational Root Theorem
According to the Rational Root Theorem, any rational root \( \frac{p}{q} \) of the polynomial equation must have \( p \) as a divisor of the constant term (189) and \( q \) as a divisor of the leading coefficient (which is 1 in this case). Since \( q = 1 \), any rational root must also be an integer.

### Step 4: Test the integer factors
We will test the integer factors of 189 to see if any of them are roots of the equation:

1. **Testing \( x = 1 \)**:
\[ 1^3 - 13(1^2) + 15(1) + 189 = 1 - 13 + 15 + 189 = 192 \quad (\text{not a root}) \]

2. **Testing \( x = -1 \)**:
\[ (-1)^3 - 13(-1)^2 + 15(-1) + 189 = -1 - 13 - 15 + 189 = 160 \quad (\text{not a root}) \]

3. **Testing \( x = 3 \)**:
\[ 3^3 - 13(3^2) + 15(3) + 189 = 27 - 117 + 45 + 189 = 144 \quad (\text{not a root}) \]

4. **Testing \( x = -3 \)**:
\[ (-3)^3 - 13(-3)^2 + 15(-3) + 189 = -27 - 117 - 45 + 189 = 0 \quad (\text{is a root}) \]

Since \( x = -3 \) is a root, we can factor the polynomial.

### Step 5: Factor the polynomial
We can factor out \( (x + 3) \) from the polynomial. Using synthetic division or polynomial long division:

\[ x^3 - 13x^2 + 15x + 189 = (x + 3)(x^2 - 16x + 63) \]

### Step 6: Factor the quadratic
Now, we need to factor the quadratic \( x^2 - 16x + 63 \):

To factor, we look for two numbers that multiply to 63 and add to -16. These numbers are -7 and -9.

Thus, we can write:

\[ x^2 - 16x + 63 = (x - 7)(x - 9) \]

### Step 7: Write the complete factorization
The complete factorization of the original polynomial is:

\[ (x + 3)(x - 7)(x - 9) = 0 \]

### Step 8: Find the roots
Setting each factor to zero gives us the roots:

1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x - 7 = 0 \) → \( x = 7 \)
3. \( x - 9 = 0 \) → \( x = 9 \)

### Conclusion
The roots of the equation \( x^3 - 13x^2 + 15x + 189 = 0 \) are \( -3, 7, \) and \( 9 \). Since we found at least one integral root, the statement in the question holds true.