To find all the zeros of the function $f(x) = x^3 + 2x^2 - 15x - 36$, we need to solve the equation $f(x) = 0$.

Step-by-Step Solution:

1. Identify Potential Rational Roots: Using the Rational Root Theorem, the potential rational roots are factors of the constant term $-36$ divided by factors of the leading coefficient $1$. This gives potential roots: $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 9, \pm 12, \pm 18, \pm 36$.

2. Test Roots Using Synthetic Division:

   • Testing $x = 3$:
     • Substitute $x = 3$ into $f(x)$: $f(3) = 3^3 + 2(3)^2 - 15(3) - 36 = 27 + 18 - 45 - 36 = 0$
     • Since $f(3) = 0$, $x = 3$ is a root.

3. Factor Out $(x - 3)$ from $f(x)$: Use synthetic division to divide $f(x)$ by $x - 3$: $x^3 + 2x^2 - 15x - 36 = (x - 3)(x^2 + 5x + 12)$

4. Solve the Quadratic $x^2 + 5x + 12 = 0$: Use the quadratic formula: $x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} = \frac{-5 \pm \sqrt{25 - 48}}{2} = \frac{-5 \pm \sqrt{-23}}{2}$ Since the discriminant is negative, this quadratic has complex roots: $x = \frac{-5 \pm \sqrt{23}i}{2}$

Final Answer:

The zeros of $f(x) = x^3 + 2x^2 - 15x - 36$ are: $x = 3, \quad x = \frac{-5 + \sqrt{23}i}{2}, \quad x = \frac{-5 - \sqrt{23}i}{2}$

Would you like a more detailed explanation of any step?

Related Questions:

1. How does the Rational Root Theorem work?
2. Why do complex roots always come in conjugate pairs?
3. Can you use synthetic division for polynomials of higher degrees?
4. How does the quadratic formula handle complex solutions?
5. What is the process to find all roots of a polynomial?

Tip:

Always start by testing small integers with the Rational Root Theorem, as they are often the simplest to verify.