Rational Zero Theorem

Finding Rational Roots of Polynomials

After learning about the Factor Theorem, we know that finding a factor $(x-c)$ is the same as finding a zero (root) $c$ of the polynomial $P(x)$. But how do we find the value of $c$, especially if the polynomial has a high degree?

Trying out all numbers is certainly not efficient. This is where the Rational Zero Theorem (or Rational Root Theorem) comes into play. This theorem helps us narrow down the list of possible rational roots of a polynomial.

Rational Zero Theorem

Let $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ be a polynomial where all coefficients ($a_n, a_{n-1}, \dots, a_1, a_0$) are integers, with $a_n \neq 0$ and $a_0 \neq 0$.

If the polynomial $P(x)$ has a rational zero (root) of the form $\frac{p}{q}$ (where $p$ and $q$ are integers, $q \neq 0$, and $\frac{p}{q}$ is a fraction in simplest form), then:

• $p$ must be a factor of the constant term $a_0$.
• $q$ must be a factor of the leading coefficient $a_n$.

This theorem only provides a list of possible rational roots. Not all values of $\frac{p}{q}$ from the list are necessarily actual roots of the polynomial. We still need to test them.

Steps for Using the Rational Zero Theorem

Here are the steps to find rational roots using this theorem, often combined with the Factor Theorem:

1. Identify Coefficients: Ensure all coefficients ($a_n, \dots, a_0$) are integers. Identify the constant term $a_0$ and the leading coefficient $a_n$.
2. List Factors of $p$: List all integer factors (positive and negative) of the constant term $a_0$.
3. List Factors of $q$: List all integer factors (positive and negative) of the leading coefficient $a_n$.
4. List Possible Roots $\frac{p}{q}$: List all possible values of $\frac{p}{q}$ by dividing each factor $p$ by each factor $q$. Simplify the fractions and remove duplicates.
5. Test Possible Roots: Test each value $\frac{p}{q}$ from the list by substituting it into $P(x)$ (using the Remainder Theorem) or using Horner's method. If the result is $P(\frac{p}{q}) = 0$, then $\frac{p}{q}$ is a rational root, and $(x - \frac{p}{q})$ (or the form $(qx - p)$) is a factor (Factor Theorem).
6. Factor Further: After finding one rational root $c$, use the quotient from Horner's method to find the remaining roots from the lower-degree polynomial.

Using the Factor Theorem and Rational Zero Theorem

Factor the polynomial $P(x) = x^3 + 2x^2 - 9x - 18$ completely.

1. Identify Coefficients:

   The coefficients are integers. $a_0 = -18$ and $a_n = 1$.

2. Factors of $p$ (from $a_0 = -18$):

   $$\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18$$

3. Factors of $q$ (from $a_n = 1$):

   $$\pm 1$$

4. Possible Roots $\frac{p}{q}$:

   Dividing all $p$ by $q = \pm 1$ yields:

   $$\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18$$

5. Test Possible Roots: Let's test some values from the list.

   • Try $x = 1$:

     $$P(1) = 1 + 2 - 9 - 18 = -24 \neq 0$$

   • Try $x = -1$:

     $$P(-1) = -1 + 2 + 9 - 18 = -8 \neq 0$$

   • Try $x = 2$:

     $$P(2) = 8 + 8 - 18 - 18 = -20 \neq 0$$

   • Try $x = -2$:

     $$P(-2) = -8 + 8 + 18 - 18 = 0$$
     .

     Success! So, $x = -2$ is a root, and $(x+2)$ is a factor.

   • Alternatively, try $x = 3$:

     $$P(3) = (3)^3 + 2(3)^2 - 9(3) - 18 = 27 + 18 - 27 - 18 = 0$$

     Success! So, $x = 3$ is a root, and $(x-3)$ is a factor.

6. Factor Further (using the root $x = 3$):

   Divide $P(x)$ by $(x-3)$ using Horner's ($c = 3$).

   $$\begin{array}{c|cccc} 3 & 1 & 2 & -9 & -18 \\ & & 3 & 15 & 18 \\ \hline & 1 & 5 & 6 & \boxed{0} \\ \end{array}$$

   The quotient is $H(x) = x^2 + 5x + 6$.

   Thus, $P(x) = (x-3)(x^2 + 5x + 6)$.

7. Factor the Quotient:

   Factor $x^2 + 5x + 6$.

   $$x^2 + 5x + 6 = (x+2)(x+3)$$

8. Complete Factorization:

   $$P(x) = (x-3)(x+2)(x+3)$$

Exercise

Factor $P(x) = 2x^3 - 3x^2 - 12x + 20$ completely using the Rational Zero Theorem and the Factor Theorem.

Answer Key

1. Identify Coefficients: $a_0 = 20$, $a_n = 2$.

2. Factors of $p$ (from 20): $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$.

3. Factors of $q$ (from 2): $\pm 1, \pm 2$.

4. Possible Roots $\frac{p}{q}$: $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20, \pm 1/2, \pm 5/2$.

5. Test Roots:

   Try $x = 2$.

   $$P(2) = 2(2)^3 - 3(2)^2 - 12(2) + 20$$

   $$P(2) = 2(8) - 3(4) - 24 + 20$$

   $$P(2) = 16 - 12 - 24 + 20$$

   $$P(2) = 4 - 4 = 0$$

   Since $P(2)=0$, $x=2$ is a root and $(x-2)$ is a factor.

6. Divide using Horner ($c = 2$):

   $$\begin{array}{c|cccc} 2 & 2 & -3 & -12 & 20 \\ & & 4 & 2 & -20 \\ \hline & 2 & 1 & -10 & \boxed{0} \\ \end{array}$$

   The quotient is $H(x) = 2x^2 + x - 10$.

   $P(x) = (x-2)(2x^2 + x - 10)$.

7. Factor the Quotient:

   Factor $2x^2 + x - 10$.

   $$2x^2 + x - 10 = (2x+5)(x-2)$$

8. Complete Factorization:

   $$P(x) = (x-2)(2x+5)(x-2)$$

   $$P(x) = (x-2)^2 (2x+5)$$