Question

Find a degree 3 polynomial with real coefficients having zeros 2 and 3 - 3i 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, we start with the given zeros: \(2\) and \(3 - 3i\). Since the coefficients of the polynomial must be real, the complex conjugate of \(3 - 3i\), which is \(3 + 3i\), must also be a zero of the polynomial.

Thus, the zeros of the polynomial are:
1. \(2\)
2. \(3 - 3i\)
3. \(3 + 3i\)

We can express the polynomial \(P(x)\) in terms of its zeros:

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

Now, let's simplify \( (x - (3 - 3i))(x - (3 + 3i)) \) first. This is a product of two conjugates:

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

Using the difference of squares, this can be written as:

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

Next, we expand \((x - 3)^2\):

\[
(x - 3)^2 = x^2 - 6x + 9
\]

Putting it all together, we have:

\[
(x - 3)^2 + 9 = x^2 - 6x + 9 + 9 = x^2 - 6x + 18
\]

Now substituting this into the polynomial gives:

\[
P(x) = (x - 2)(x^2 - 6x + 18)
\]

Now, we expand this product:

\[
P(x) = x(x^2 - 6x + 18) - 2(x^2 - 6x + 18)
\]
\[
= x^3 - 6x^2 + 18x - 2x^2 + 12x - 36
\]
\[
= x^3 - 8x^2 + 30x - 36
\]

Thus, the final polynomial is:

\[
P(x) = x^3 - 8x^2 + 30x - 36
\]

Putting it all together, the full equation is:

\[
\boxed{P(x) = x^3 - 8x^2 + 30x - 36}
\]