To find the zeros of the cubic function $f(x) = 4x^3 + 3x^2 + 36x + 27$, we need to solve the equation:

$4x^3 + 3x^2 + 36x + 27 = 0$

Step 1: Look for Rational Roots

We can use the Rational Root Theorem to test possible rational roots. The Rational Root Theorem suggests that any rational solution is of the form:

$\frac{p}{q}$

Where $p$ is a factor of the constant term (27), and $q$ is a factor of the leading coefficient (4). The factors of 27 are $\pm 1, \pm 3, \pm 9, \pm 27$, and the factors of 4 are $\pm 1, \pm 2, \pm 4$.

Thus, the possible rational roots are:

$\pm 1, \pm \frac{1}{2}, \pm 3, \pm \frac{3}{2}, \pm 9, \pm \frac{9}{2}, \pm 27, \pm \frac{27}{2}$

Step 2: Test Rational Roots

Let's begin by testing $x = -3$:

$f(-3) = 4(-3)^3 + 3(-3)^2 + 36(-3) + 27 = -108 + 27 - 108 + 27 = 0$

Since $f(-3) = 0$, $x = -3$ is a root.

Step 3: Polynomial Division

Now that we know $x = -3$ is a root, we can divide the cubic polynomial $f(x) = 4x^3 + 3x^2 + 36x + 27$ by $x + 3$ using synthetic division or long division.

Dividing $4x^3 + 3x^2 + 36x + 27$ by $x + 3$, we get:

$4x^3 + 3x^2 + 36x + 27 = (x + 3)(4x^2 - 9x + 9)$

Step 4: Solve the Quadratic Equation

Next, we solve the quadratic equation $4x^2 - 9x + 9 = 0$. Using the quadratic formula:

$x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(4)(9)}}{2(4)}$

$x = \frac{9 \pm \sqrt{81 - 144}}{8}$

$x = \frac{9 \pm \sqrt{-63}}{8}$

$x = \frac{9 \pm i\sqrt{63}}{8}$

$x = \frac{9 \pm i\sqrt{9 \times 7}}{8}$

$x = \frac{9 \pm 3i\sqrt{7}}{8}$

Thus, the roots of the equation are:

$x = -3, \quad x = \frac{9 + 3i\sqrt{7}}{8}, \quad x = \frac{9 - 3i\sqrt{7}}{8}$

Final Answer:

The zeros of $f(x) = 4x^3 + 3x^2 + 36x + 27$ are:

$x = -3, \quad x = \frac{9 + 3i\sqrt{7}}{8}, \quad x = \frac{9 - 3i\sqrt{7}}{8}$

Would you like further details on any of these steps? Or do you have any other questions?

Here are some related questions:

1. How do you use synthetic division to divide polynomials?
2. What is the Rational Root Theorem, and how does it work?
3. How do you solve a cubic equation by factoring?
4. What is the significance of imaginary roots in real-world applications?
5. How do you apply the quadratic formula to find complex roots?

Tip: Always check if a root can be easily found using synthetic division before resorting to the quadratic formula for the remaining factor.