Question

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Part 1
Find an​ nth-degree polynomial function with real coefficients satisfying the given conditions. If you are using a graphing​ utility, use it to graph the function and verify the real zeros and the given function value.
nequals​3;
4 and 5 i are​ zeros;
f left parenthesis negative 1 right parenthesis equals negative 260
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Part 1
f left parenthesis x right parenthesisequals
  
enter your response here
​(Type an expression using x as the variable. Simplify your​ answer.)

Answer 1

To construct an \( n \)-th degree polynomial function with real coefficients given the conditions that \( n = 3 \), and the zeros are \( 4 \) and \( 5i \), we proceed as follows:

1. **Identify the Zeros**:
- Zeros: \( 4 \), \( 5i \)
- Since the polynomial has real coefficients, the complex conjugate of \( 5i \), which is \( -5i \), must also be a zero.

2. **Form the Factors**:
- The polynomial can be expressed as:
\[
f(x) = (x - 4)(x - 5i)(x + 5i)
\]

3. **Simplify the Complex Factors**:
\[
(x - 5i)(x + 5i) = x^2 + 25
\]
Thus, the polynomial becomes:
\[
f(x) = (x - 4)(x^2 + 25)
\]

4. **Expand the Polynomial**:
\[
f(x) = (x - 4)(x^2 + 25) = x(x^2 + 25) - 4(x^2 + 25) = x^3 + 25x - 4x^2 - 100
\]
Rearranging gives:
\[
f(x) = x^3 - 4x^2 + 25x - 100
\]

5. **Use the Given Condition**:
We need \( f(-1) = -260 \):
\[
f(-1) = (-1)^3 - 4(-1)^2 + 25(-1) - 100 = -1 - 4 - 25 - 100 = -130
\]
To satisfy the condition, we scale the polynomial. Let's denote \( f(x) = k(x^3 - 4x^2 + 25x - 100) \) and find \( k \) such that:
\[
k(-130) = -260 \implies k = \frac{-260}{-130} = 2
\]

6. **Final Polynomial**:
Therefore, the polynomial function is:
\[
f(x) = 2(x^3 - 4x^2 + 25x - 100)
\]
Expanding this results in:
\[
f(x) = 2x^3 - 8x^2 + 50x - 200
\]

Hence, the required \( 3 \)-degree polynomial function is:
\[
\boxed{2x^3 - 8x^2 + 50x - 200}
\]