Question
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.
nequals3;
3 and 4 i are zeros;
f left parenthesis 1 right parenthesis equals negative 102
Question content area bottom
Part 1
f left parenthesis x right parenthesisequals
enter your response here
(Type an expression using x as the variable. Simplify your answer.)
nequals3;
3 and 4 i are zeros;
f left parenthesis 1 right parenthesis equals negative 102
Question content area bottom
Part 1
f left parenthesis x right parenthesisequals
enter your response here
(Type an expression using x as the variable. Simplify your answer.)
Answers
GPT-4o mini
To construct a polynomial function of degree 3 with the real coefficients and given zeros, we start by recognizing the zeros provided: \(3\) and \(4i\). Since the coefficients are real, the complex conjugate \(4i\) also indicates that \(-4i\) is a zero. Therefore, the polynomial will have the following zeros:
1. \(3\)
2. \(4i\)
3. \(-4i\)
### Step 1: Construct the polynomial from its zeros
The polynomial \(f(x)\) can be expressed in factored form based on its zeros:
\[
f(x) = (x - 3)(x - 4i)(x + 4i)
\]
Notice that \( (x - 4i)(x + 4i) \) can be simplified using the difference of squares:
\[
(x - 4i)(x + 4i) = x^2 + 16
\]
Thus, the polynomial becomes:
\[
f(x) = (x - 3)(x^2 + 16)
\]
### Step 2: Expand the polynomial
Now, we will expand the expression:
\[
f(x) = (x - 3)(x^2 + 16)
\]
Distributing yields:
\[
f(x) = x(x^2 + 16) - 3(x^2 + 16)
\]
Breaking it down:
1. \(x(x^2 + 16) = x^3 + 16x\)
2. \(-3(x^2 + 16) = -3x^2 - 48\)
Combining these results gives:
\[
f(x) = x^3 + 16x - 3x^2 - 48
\]
Rearranging the terms in descending order:
\[
f(x) = x^3 - 3x^2 + 16x - 48
\]
### Step 3: Adjust the function to satisfy \(f(1) = -102\)
Now, we need to ensure \(f(1) = -102\). We first calculate \(f(1)\):
\[
f(1) = 1^3 - 3(1)^2 + 16(1) - 48 = 1 - 3 + 16 - 48 = -34
\]
Since \(f(1) = -34\), we need to find a scalar \(k\) so that:
\[
k \cdot (-34) = -102 \implies k = \frac{-102}{-34} = 3
\]
### Final Polynomial
Thus, we multiply the polynomial by \(3\):
\[
f(x) = 3(x^3 - 3x^2 + 16x - 48)
\]
Expanding gives:
\[
f(x) = 3x^3 - 9x^2 + 48x - 144
\]
### Final Answer
The nth-degree polynomial function satisfying all given conditions is:
\[
\boxed{3x^3 - 9x^2 + 48x - 144}
\]
1. \(3\)
2. \(4i\)
3. \(-4i\)
### Step 1: Construct the polynomial from its zeros
The polynomial \(f(x)\) can be expressed in factored form based on its zeros:
\[
f(x) = (x - 3)(x - 4i)(x + 4i)
\]
Notice that \( (x - 4i)(x + 4i) \) can be simplified using the difference of squares:
\[
(x - 4i)(x + 4i) = x^2 + 16
\]
Thus, the polynomial becomes:
\[
f(x) = (x - 3)(x^2 + 16)
\]
### Step 2: Expand the polynomial
Now, we will expand the expression:
\[
f(x) = (x - 3)(x^2 + 16)
\]
Distributing yields:
\[
f(x) = x(x^2 + 16) - 3(x^2 + 16)
\]
Breaking it down:
1. \(x(x^2 + 16) = x^3 + 16x\)
2. \(-3(x^2 + 16) = -3x^2 - 48\)
Combining these results gives:
\[
f(x) = x^3 + 16x - 3x^2 - 48
\]
Rearranging the terms in descending order:
\[
f(x) = x^3 - 3x^2 + 16x - 48
\]
### Step 3: Adjust the function to satisfy \(f(1) = -102\)
Now, we need to ensure \(f(1) = -102\). We first calculate \(f(1)\):
\[
f(1) = 1^3 - 3(1)^2 + 16(1) - 48 = 1 - 3 + 16 - 48 = -34
\]
Since \(f(1) = -34\), we need to find a scalar \(k\) so that:
\[
k \cdot (-34) = -102 \implies k = \frac{-102}{-34} = 3
\]
### Final Polynomial
Thus, we multiply the polynomial by \(3\):
\[
f(x) = 3(x^3 - 3x^2 + 16x - 48)
\]
Expanding gives:
\[
f(x) = 3x^3 - 9x^2 + 48x - 144
\]
### Final Answer
The nth-degree polynomial function satisfying all given conditions is:
\[
\boxed{3x^3 - 9x^2 + 48x - 144}
\]