Question

Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions.
nequals​4;
2 i and 3 i are​ zeros;
f left parenthesis negative 2 right parenthesisequals104
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Part 1
​f(x)equals
  
enter your response here
​(Type an expression using x as the variable. Simplify your​ answer.)

Answer 1

Given that the degree of the polynomial is \( n = 4 \), and that \( 2i \) and \( 3i \) are zeros, we must also include their conjugate pairs due to the requirement for real coefficients. Therefore, the zeros of the polynomial are:

- \( 2i \) and \( -2i \)
- \( 3i \) and \( -3i \)

This means we can express the polynomial in factored form as:

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

First, we can simplify the factors involving the complex zeros:

\[
(x - 2i)(x + 2i) = x^2 + 4
\]

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

Thus, the polynomial function can be rewritten as:

\[
f(x) = k(x^2 + 4)(x^2 + 9)
\]

Next, we expand the polynomial:

\[
f(x) = k((x^2)(x^2) + 9x^2 + 4x^2 + 36) = k(x^4 + 13x^2 + 36)
\]

Now, we have:

\[
f(x) = k(x^4 + 13x^2 + 36)
\]

We need to determine the value of \( k \) so that \( f(-2) = 104 \):

\[
f(-2) = k((-2)^4 + 13(-2)^2 + 36)
\]

Calculating \( f(-2) \):

\[
(-2)^4 = 16, \quad 13(-2)^2 = 13 \cdot 4 = 52
\]

Thus:

\[
f(-2) = k(16 + 52 + 36) = k(104)
\]

To satisfy \( f(-2) = 104 \):

\[
k(104) = 104
\]

This gives us:

\[
k = 1
\]

Therefore, the polynomial function is:

\[
f(x) = 1(x^4 + 13x^2 + 36) = x^4 + 13x^2 + 36
\]

The final answer is:

\[
\boxed{x^4 + 13x^2 + 36}
\]