Find a degree 3 polynomial with a leading coefficient of 1, and zeros -1,
− 3 i , and
3 i . Simplify your answer so that it has only real numbers as coefficients. Be sure to write the full equation, including
P ( x )

To find a degree 3 polynomial with a leading coefficient of 1 and zeros at $-1$, $-3i$, and $3i$, we follow these steps:

Step 1: Form the factors from the zeros

For each zero $r$, the corresponding factor of the polynomial is $(x - r)$.

• For $-1$, the factor is $(x + 1)$.
• For $-3i$, the factor is $(x + 3i)$.
• For $3i$, the factor is $(x - 3i)$.

Thus, the polynomial can be expressed as: $P(x) = (x + 1)(x + 3i)(x - 3i)$

Step 2: Multiply the complex conjugate factors

First, simplify $(x + 3i)(x - 3i)$. This is a difference of squares: $(x + 3i)(x - 3i) = x^2 - (3i)^2 = x^2 - (-9) = x^2 + 9$

Step 3: Multiply the remaining factor

Now, multiply the result from Step 2 by $(x + 1)$: $P(x) = (x + 1)(x^2 + 9)$

Distribute $(x + 1)$ to $x^2 + 9$: $P(x) = x(x^2 + 9) + 1(x^2 + 9)$ $P(x) = x^3 + 9x + x^2 + 9$

Step 4: Write the final simplified polynomial

Rearrange the terms in descending order: $P(x) = x^3 + x^2 + 9x + 9$

Final Answer:

$P(x) = x^3 + x^2 + 9x + 9$

This is the degree 3 polynomial with the given zeros and a leading coefficient of 1.

Would you like further details or explanations? Here are 5 related questions:

1. How do you identify the factors of a polynomial from its zeros?
2. Why is it necessary to multiply conjugate pairs to get real coefficients?
3. What are complex conjugates and why do they matter in polynomials with real coefficients?
4. Can you explain how the difference of squares formula works in this context?
5. What other methods can be used to find polynomials from given zeros?

Tip: When a polynomial has complex roots, they must occur in conjugate pairs for the polynomial to have real coefficients.