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To solve this problem, we can use the **Rational Zeros Theorem**, which states that if a polynomial has rational zeros, then they are in the form \( \pm \frac{p}{q} \), where:

- \( p \) is a factor of the constant term.
- \( q \) is a factor of the leading coefficient.

The polynomial given is:

\[
h(x) = 6x^3 + 31x^2 + 4x - 5
\]

### Step 1: Identify Possible Rational Zeros
- The constant term is \(-5\), so its factors (\( p \)) are \( \pm 1, \pm 5 \).
- The leading coefficient is \( 6 \), so its factors (\( q \)) are \( \pm 1, \pm 2, \pm 3, \pm 6 \).

Using the Rational Zeros Theorem, the possible rational zeros are:

\[
\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{3}, \pm \frac{5}{6}
\]

### Step 2: Test Possible Zeros Using Synthetic Division
We will test these potential zeros by plugging them into the polynomial to find which one(s) result in \( h(x) = 0 \). A common approach is to start with the simplest candidates, such as \( \pm 1 \).

1. **Test \( x = 1 \)**:
   \[
   h(1) = 6(1)^3 + 31(1)^2 + 4(1) - 5 = 6 + 31 + 4 - 5 = 36 \quad (\neq 0)
   \]

2. **Test \( x = -1 \)**:
   \[
   h(-1) = 6(-1)^3 + 31(-1)^2 + 4(-1) - 5 = -6 + 31 - 4 - 5 = 16 \quad (\neq 0)
   \]

3. **Test \( x = 5 \)**:
   \[
   h(5) = 6(5)^3 + 31(5)^2 + 4(5) - 5 = 6(125) + 31(25) + 20 - 5 = 750 + 775 + 20 - 5 = 1540 \quad (\neq 0)
   \]

4. **Test \( x = -5 \)**:
   \[
   h(-5) = 6(-5)^3 + 31(-5)^2 + 4(-5) - 5 = 6(-125) + 31(25) - 20 - 5 = -750 + 775 - 20 - 5 = 0
   \]

Since \( h(-5) = 0 \), \( x = -5 \) is a zero.

### Step 3: Perform Polynomial Division
Now that \( x = -5 \) is a zero, we divide \( h(x) \) by \( x + 5 \). We can use synthetic division for this:

- **Perform synthetic division**:

|   | -5 | |
|---|----|---|
| 6 | 31 | 4 | -5 |
|   |    | -30 | -5 | 45 |
|----------------------|
| 6 | 1 | -1 | 0 |

The quotient is \( 6x^2 + x - 1 \).

### Step 4: Factor the Quotient
We now have:

\[
h(x) = (x + 5)(6x^2 + x - 1)
\]

To find the remaining zeros, we solve \( 6x^2 + x - 1 = 0 \). We can factor this quadratic equation by finding two numbers that multiply to \( -6 \) and add to \( 1 \). Factoring yields:

\[
6x^2 + x - 1 = (3x - 1)(2x + 1)
\]

### Step 5: Find All Zeros
Setting each factor equal to zero:

1. \( x + 5 = 0 \) gives \( x = -5 \).
2. \( 3x - 1 = 0 \) gives \( x = \frac{1}{3} \).
3. \( 2x + 1 = 0 \) gives \( x = -\frac{1}{2} \).

### Final Answer
The zeros of \( h(x) \) are:

\[
x = -5, \quad x = \frac{1}{3}, \quad x = -\frac{1}{2}
\]

Would you like more details, or have any questions on this? Here are five related questions for further understanding:

1. What is the Rational Zeros Theorem, and how does it help in finding zeros of a polynomial?
2. Can all polynomials have rational zeros? Why or why not?
3. How is synthetic division performed, and why is it used in this context?
4. What should be done if a polynomial has no rational zeros but still needs all zeros found?
5. What are the steps to factor a quadratic expression, and when is it necessary?

**Tip:** When using the Rational Zeros Theorem, always start with simpler factors like \( \pm 1 \) to quickly find rational zeros before testing more complex possibilities.

The function below has at least one rational zero. Use this fact to find all zeros of the function. h(x) = 6x^3 + 31x^2 + 4x - 5

Learn how to find all zeros of the polynomial h(x) = 6x^3 + 31x^2 + 4x - 5 using the Rational Zeros Theorem and synthetic division. This example explores polynomial factorization and solving for exact values, suitable for high school students in Grades 10-12.

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