find the rational zeros of f(x)=3x^3 −19x^2 +33x−9

To find the rational zeros of the polynomial function \( f(x) = 3x^3 - 19x^2 + 33x - 9 \), we can use the **Rational Root Theorem**. This theorem provides a systematic way to test potential rational zeros by considering the factors of the constant term and the leading coefficient.

### Steps:

1. **Identify the coefficients:**
   - The leading coefficient (coefficient of \(x^3\)) is \(3\).
   - The constant term is \(-9\).

2. **List the factors:**
   - Factors of the constant term \(-9\): \(\pm 1, \pm 3, \pm 9\).
   - Factors of the leading coefficient \(3\): \(\pm 1, \pm 3\).

3. **Form possible rational zeros:**
   The possible rational zeros are the factors of the constant term divided by the factors of the leading coefficient:
   \[
   \text{Possible rational zeros} = \pm \left\{ \frac{1}{1}, \frac{3}{1}, \frac{9}{1}, \frac{1}{3}, \frac{3}{3}, \frac{9}{3} \right\}
   \]
   Simplifying, we get:
   \[
   \text{Possible rational zeros} = \pm 1, \pm 3, \pm 9, \pm \frac{1}{3}
   \]

4. **Test these values using synthetic division or substitution:**
   Now, we'll test these values in the function \( f(x) \).

   - For \( x = 1 \):
     \[
     f(1) = 3(1)^3 - 19(1)^2 + 33(1) - 9 = 3 - 19 + 33 - 9 = 8 \quad (\text{Not a zero})
     \]

   - For \( x = -1 \):
     \[
     f(-1) = 3(-1)^3 - 19(-1)^2 + 33(-1) - 9 = -3 - 19 - 33 - 9 = -64 \quad (\text{Not a zero})
     \]

   - For \( x = 3 \):
     \[
     f(3) = 3(3)^3 - 19(3)^2 + 33(3) - 9 = 81 - 171 + 99 - 9 = 0 \quad (\text{Zero found})
     \]
     So, \( x = 3 \) is a rational zero.

5. **Factor the polynomial using synthetic division:**
   Now that we know \( x = 3 \) is a zero, we can factor the polynomial using synthetic division. Dividing \( f(x) \) by \( (x - 3) \):

   \[
   \begin{array}{r|rrrr}
   3 & 3 & -19 & 33 & -9 \\
     &   & 9  & -30 & 9 \\
   \hline
     & 3 & -10 & 3  & 0 \\
   \end{array}
   \]
   The quotient is \( 3x^2 - 10x + 3 \), so:
   \[
   f(x) = (x - 3)(3x^2 - 10x + 3)
   \]

6. **Solve the quadratic equation:**
   To find the remaining zeros, solve \( 3x^2 - 10x + 3 = 0 \) using the quadratic formula:
   \[
   x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(3)}}{2(3)} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6}
   \]
   So, the solutions are:
   \[
   x = \frac{10 + 8}{6} = \frac{18}{6} = 3 \quad \text{and} \quad x = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3}
   \]

### Rational Zeros:
Thus, the rational zeros of \( f(x) = 3x^3 - 19x^2 + 33x - 9 \) are:
\[
x = 3 \quad \text{(repeated root)}, \quad x = \frac{1}{3}
\]

Would you like more details on any step or have further questions?

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Here are 5 related questions:
1. How does synthetic division work step by step?
2. Can you explain how the Rational Root Theorem applies to other types of polynomials?
3. What happens if none of the potential rational zeros work?
4. How can we factor quadratic equations more easily?
5. What is the significance of repeated roots in polynomials?

**Tip:** When using the Rational Root Theorem, always test the simpler values first, as they are more likely to be roots.

find the rational zeros of f(x) = 3x^3 − 19x^2 + 33x − 9

Discover how to find the rational zeros of the polynomial function f(x) = 3x^3 − 19x^2 + 33x − 9 using the Rational Root Theorem and synthetic division. This step-by-step guide walks through testing possible rational zeros and solving the resulting quadratic equation. Suitable for students in Grades 10-12.

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