We are given the polynomial:

$f(x) = x^3 + 4x^2 - 4x - 16$

To find the zeroes of this polynomial, we can use the Rational Root Theorem to suggest possible rational roots. The possible rational roots are factors of the constant term (-16) divided by the factors of the leading coefficient (1).

The factors of -16 are: $\pm 1, \pm 2, \pm 4, \pm 8, \pm 16$

The factors of 1 (leading coefficient) are: $\pm 1$

Thus, the possible rational roots are: $\pm 1, \pm 2, \pm 4, \pm 8, \pm 16$

Step 1: Testing potential roots

We test $x = -2$ first by substituting it into the polynomial:

$f(-2) = (-2)^3 + 4(-2)^2 - 4(-2) - 16 = -8 + 16 + 8 - 16 = 0$

So, $x = -2$ is a root.

Step 2: Polynomial Division

Now that we know $x + 2$ is a factor of $f(x)$, we can divide $f(x)$ by $x + 2$ using synthetic or long division.

Dividing $f(x)$ by $x + 2$:

$f(x) = (x + 2)(x^2 + 2x - 8)$

Step 3: Finding the zeroes of the quadratic factor

Next, solve $x^2 + 2x - 8 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 2$, and $c = -8$. Substituting into the formula:

$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-8)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2}$

$x = \frac{-2 \pm 6}{2}$

This gives two solutions: $x = \frac{-2 + 6}{2} = 2 \quad \text{or} \quad x = \frac{-2 - 6}{2} = -4$

Final Answer:

The zeroes of the polynomial are: $x = -2, 2, -4$

Let me know if you'd like further details or have any questions!

Related Questions:

1. How would you use synthetic division to divide polynomials?
2. What is the Rational Root Theorem and how does it help in solving polynomial equations?
3. How do you solve a cubic equation by factoring?
4. What is the discriminant in the quadratic formula and how does it affect the number of real solutions?
5. How can you verify your roots by substituting them back into the original polynomial?

Tip: If you're unsure about which roots to try, always start with small numbers like 1, -1, 2, or -2, which are more likely to give integer solutions.