Find an [tex]\(n\)[/tex]-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.

[tex]\[
\begin{array}{l}
n = 4 \\
-1, 3, \text{ and } 2 + 3i \text{ are zeros} \\
f(1) = -120
\end{array}
\][/tex]



Answer :

To find a 4th-degree polynomial with the given conditions, we need to follow these steps:

### Step 1: Identify the Polynomial from its Roots
Given the roots are [tex]\(-1\)[/tex], [tex]\(3\)[/tex], [tex]\(2+3i\)[/tex], and since complex roots always come in conjugate pairs for polynomials with real coefficients, the other root must be [tex]\(2-3i\)[/tex].

The polynomial can be constructed by multiplying the factors corresponding to its roots:
[tex]\[ f(x) = a(x + 1)(x - 3)(x - (2 + 3i))(x - (2 - 3i)) \][/tex]

### Step 2: Expand the Polynomial
First, expand the complex conjugate factors:
[tex]\[ (x - (2 + 3i))(x - (2 - 3i)) \][/tex]

Using the fact that [tex]\((x - (a + bi))(x - (a - bi)) = (x - a)^2 + b^2\)[/tex]:
[tex]\[ (x - 2)^2 + (3i)^2 = (x - 2)^2 + 9 \][/tex]
[tex]\[ (x^2 - 4x + 4) + 9 = x^2 - 4x + 13 \][/tex]

Next, include the other factors [tex]\((x + 1)\)[/tex] and [tex]\((x - 3)\)[/tex]:
[tex]\[ f(x) = a(x + 1)(x - 3)(x^2 - 4x + 13) \][/tex]

Now, expand this expression:
First multiply [tex]\((x + 1)(x - 3)\)[/tex]:
[tex]\[ (x + 1)(x - 3) = x^2 - 3x + x - 3 = x^2 - 2x - 3 \][/tex]

Now multiply this result by [tex]\((x^2 - 4x + 13)\)[/tex]:
[tex]\[ f(x) = a(x^2 - 2x - 3)(x^2 - 4x + 13) \][/tex]

Expanding this:
[tex]\[ (x^2 - 2x - 3)(x^2 - 4x + 13) = x^4 - 4x^3 + 13x^2 - 2x^3 + 8x^2 - 26x - 3x^2 + 12x - 39 \][/tex]

Combine like terms:
[tex]\[ x^4 - 6x^3 + 18x^2 - 14x - 39 \][/tex]

Thus, the polynomial becomes:
[tex]\[ f(x) = a(x^4 - 6x^3 + 18x^2 - 14x - 39) \][/tex]

### Step 3: Determine the Scaling Factor [tex]\(a\)[/tex]
We are given [tex]\(f(1) = -120\)[/tex], so we need to find [tex]\(a\)[/tex] such that:
[tex]\[ f(1) = a(1^4 - 6 \cdot 1^3 + 18 \cdot 1^2 - 14 \cdot 1 - 39) = -120 \][/tex]

Calculate [tex]\(f(1)\)[/tex]:
[tex]\[ 1 - 6 + 18 - 14 - 39 = -40 \][/tex]

So we have:
[tex]\[ a(-40) = -120 \][/tex]
[tex]\[ a = \frac{-120}{-40} = 3 \][/tex]

### Step 4: Write the Final Polynomial
The final polynomial is:
[tex]\[ f(x) = 3(x^4 - 6x^3 + 18x^2 - 14x - 39) \][/tex]

Expanding this:
[tex]\[ f(x) = 3x^4 - 18x^3 + 54x^2 - 42x - 117 \][/tex]

So, the polynomial function satisfying all the conditions is:
[tex]\[ f(x) = 3x^4 - 18x^3 + 54x^2 - 42x - 117 \][/tex]

### Verification
To check:
When [tex]\(x = 1\)[/tex],
[tex]\[ f(1) = 3(1^4) - 18(1^3) + 54(1^2) - 42(1) - 117 \][/tex]
[tex]\[ = 3 - 18 + 54 - 42 - 117 \][/tex]
[tex]\[ = -120 \][/tex]

Therefore, the polynomial function is verified to meet the conditions given in the problem.