The example above was for quadratics, but similar thinking expands the pool to any polynomial with integer coefficients. The numerators of rational roots would have to be $$ \vee $$, and the denominators would have to be $$ \vee $$ to form the

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It looks like you're discussing the **Rational Root Theorem**, which provides a way to identify possible rational roots of a polynomial with integer coefficients. Let's complete your explanation:

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The example above was for quadratics, but similar thinking expands the pool to **any polynomial with integer coefficients**. According to the **Rational Root Theorem**, the possible rational roots of such a polynomial can be determined as follows:

- **Numerators of rational roots** must be **divisors of the constant term**.
  
  $$
  \text{Numerators} = \text{Divisors of the constant term}
  $$
  
- **Denominators of rational roots** must be **divisors of the leading coefficient**.
  
  $$
  \text{Denominators} = \text{Divisors of the leading coefficient}
  $$
  
To form the possible rational roots, you take each divisor of the constant term and divide it by each divisor of the leading coefficient. This gives you all potential candidates for rational roots, which you can then test (e.g., via synthetic division or plugging them back into the polynomial) to determine if they are actual roots.

**Example:**

Consider the polynomial:
$$
P(x) = 2x^3 - 3x^2 - 8x + 12
$$

1. **Identify the constant term and the leading coefficient:**
   - Constant term = 12
   - Leading coefficient = 2

2. **List the divisors:**
   - Divisors of 12 (constant term): $$ \pm1, \pm2, \pm3, \pm4, \pm6, \pm12 $$
   - Divisors of 2 (leading coefficient): $$ \pm1, \pm2 $$

3. **Form possible rational roots ($$ \frac{p}{q} $$):**
   $$
   \pm1, \pm\frac{1}{2}, \pm2, \pm3, \pm\frac{3}{2}, \pm4, \pm\frac{4}{2} (\text{which simplifies to } \pm2), \pm6, \pm\frac{6}{2} (\text{which simplifies to } \pm3), \pm12, \pm\frac{12}{2} (\text{which simplifies to } \pm6)
   $$
   
   After removing duplicates:
   $$
   \pm1, \pm\frac{1}{2}, \pm2, \pm3, \pm4, \pm6, \pm12
   $$

4. **Test these possible roots in the polynomial to find actual roots.**

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By systematically applying the Rational Root Theorem, you can efficiently narrow down the list of potential rational solutions for any polynomial with integer coefficients.

If you have a specific polynomial you'd like to analyze, feel free to share it, and we can work through the possible rational roots together!
To find possible rational roots of a polynomial with integer coefficients, list all divisors of the constant term as numerators and all divisors of the leading coefficient as denominators. Then, form fractions by dividing each numerator by each denominator to get the possible rational roots.

The example above was for quadratics, but similar thinking expands the pool to any polynomial with integer coefficients. The numerators of rational roots would have to be \( \vee \), and the denominators would have to be \( \vee \) to form the

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