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:

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.