Question

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

Answer 1

To find a degree 3 polynomial with real coefficients having zeros at \(3\) and \(2i\), we first note that because the coefficients are real, the complex conjugate of \(2i\), which is \(-2i\), must also be a zero of the polynomial.

The zeros of the polynomial are:
- \(3\)
- \(2i\)
- \(-2i\)

The polynomial can be expressed in factored form as:
\[
P(x) = (x - 3)(x - 2i)(x + 2i)
\]

Next, we simplify the factors corresponding to the complex zeros:
\[
(x - 2i)(x + 2i) = x^2 + 4
\]
This uses the identity \( (a - b)(a + b) = a^2 - b^2 \) where \( a = x \) and \( b = 2i \).

Now, substituting this back into the polynomial:
\[
P(x) = (x - 3)(x^2 + 4)
\]

Next, we expand this product:
\[
P(x) = x(x^2 + 4) - 3(x^2 + 4)
\]
\[
= x^3 + 4x - 3x^2 - 12
\]
Rearranging the terms gives us the polynomial in standard form:
\[
P(x) = x^3 - 3x^2 + 4x - 12
\]

Thus, the final expression for the polynomial \( P \) is:
\[
\boxed{P(x) = x^3 - 3x^2 + 4x - 12}
\]