The problem requires finding all zeros of the polynomial function:

$f(x) = 8x^3 - 2x^2 - 15x + 9$

Steps to Solve:

1. Identify possible rational roots using the Rational Root Theorem: The possible roots are $\pm \frac{p}{q}$, where $p$ is a factor of the constant term (9) and $q$ is a factor of the leading coefficient (8). The possible rational roots are:

   $\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm \frac{3}{8}, \pm 9, \pm \frac{9}{2}, \pm \frac{9}{4}, \pm \frac{9}{8}$

2. Test rational roots: Substitute the possible roots into the function $f(x)$. After testing, we find that $x = \frac{3}{2}$ is a root.

3. Perform synthetic division: Divide $f(x)$ by $x - \frac{3}{2}$ to reduce the degree of the polynomial.

   Result of division: $f(x) = \left(x - \frac{3}{2}\right)\left(8x^2 + 10x - 6\right)$

4. Solve the quadratic $8x^2 + 10x - 6 = 0$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substituting $a = 8$, $b = 10$, and $c = -6$: $x = \frac{-10 \pm \sqrt{10^2 - 4(8)(-6)}}{2(8)}$ $x = \frac{-10 \pm \sqrt{100 + 192}}{16}$ $x = \frac{-10 \pm \sqrt{292}}{16}$ Simplify: $x = \frac{-10 \pm 2\sqrt{73}}{16} = \frac{-5 \pm \sqrt{73}}{8}$

Final Zeros:

$x = \frac{3}{2}, \quad x = \frac{-5 + \sqrt{73}}{8}, \quad x = \frac{-5 - \sqrt{73}}{8}$

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Related Questions:

1. How does the Rational Root Theorem help in solving polynomial equations?
2. What are other methods for finding roots of polynomials?
3. Why is synthetic division useful for simplifying polynomials?
4. How do irrational roots affect the solution of a polynomial equation?
5. Can complex roots occur in a cubic polynomial, and how would they be