Question

Find a degree 3 polynomial with real coefficients having zeros 2 and 3i and a lead coefficient of 1. Write P in expanded form. Be sure to write the full equation, including P(x)=.

Answer 1

Given the zeros of the polynomial \(2\) and \(3i\), and knowing that the coefficients are real, the complex conjugate \( -3i \) must also be a zero. Thus, the zeros of the polynomial are \(2\), \(3i\), and \(-3i\).

To write the polynomial, we can use the fact that a polynomial can be expressed in its factored form as follows:

\[
P(x) = (x - 2)(x - 3i)(x + 3i)
\]

Now, we can simplify \( (x - 3i)(x + 3i) \) using the difference of squares:

\[
(x - 3i)(x + 3i) = x^2 - (3i)^2 = x^2 - (-9) = x^2 + 9
\]

Thus, we can now rewrite the polynomial as:

\[
P(x) = (x - 2)(x^2 + 9)
\]

Now we can expand this expression:

\[
P(x) = x(x^2 + 9) - 2(x^2 + 9)
\]
\[
= x^3 + 9x - 2x^2 - 18
\]
\[
= x^3 - 2x^2 + 9x - 18
\]

Thus, the polynomial \( P(x) \) is:

\[
P(x) = x^3 - 2x^2 + 9x - 18
\]

So, the final answer is:

\[
P(x) = x^3 - 2x^2 + 9x - 18
\]