The polynomial given is:

$P(x) = x^3 - x^2 - 4x - 6$

with one known zero: $x = 3$.

To find the remaining zeros, let's start by dividing the polynomial $P(x)$ by $x - 3$, since $x = 3$ is a root of the polynomial. After division, the result will be a quadratic polynomial, which we can solve to find the other zeros.

Step-by-Step Solution:

1. Divide $P(x)$ by $x - 3$: Using synthetic or polynomial division, we divide $x^3 - x^2 - 4x - 6$ by $x - 3$.

2. Result of the Division: Performing the division gives us: $P(x) = (x - 3)(x^2 + 2x + 2)$

3. Solve the Quadratic Equation $x^2 + 2x + 2 = 0$: To find the remaining zeros, solve the quadratic equation: $x^2 + 2x + 2 = 0$

   Using the quadratic formula: $x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$ $x = \frac{-2 \pm \sqrt{4 - 8}}{2}$ $x = \frac{-2 \pm \sqrt{-4}}{2}$ $x = \frac{-2 \pm 2i}{2}$ $x = -1 \pm i$

4. Zeros of $P(x)$: The zeros of $P(x)$ are: $x = 3, \quad x = -1 + i, \quad \text{and} \quad x = -1 - i$

So, the remaining zeros are $x = -1 + i$ and $x = -1 - i$.

Would you like more details on any step, or do you have any questions?

Related Questions:

1. What is synthetic division, and how is it used with polynomials?
2. How do complex roots come in conjugate pairs for polynomials with real coefficients?
3. Can the quadratic formula always be used to find zeros of polynomials?
4. What are other methods to find roots of higher-degree polynomials?
5. How does knowing one zero help in factoring a polynomial?

Tip:

Always check for real and complex solutions when solving polynomials to ensure all zeros are found.