Answer :
Let's find a cubic (third-degree) polynomial function [tex]\( f(x) \)[/tex] with real coefficients given the following conditions:
1. The polynomial has real coefficients.
2. It has zeros at 3 and [tex]\( 2i \)[/tex] where [tex]\( i \)[/tex] is the imaginary unit.
3. It satisfies the condition [tex]\( f(2) = 16 \)[/tex].
### Step-by-Step Solution:
#### 1. Determine All Zeros
First, note that since the polynomial has real coefficients and [tex]\( 2i \)[/tex] is a zero, its complex conjugate [tex]\( -2i \)[/tex] must also be a zero. Therefore, the zeros of the polynomial are:
- [tex]\( 3 \)[/tex]
- [tex]\( 2i \)[/tex]
- [tex]\( -2i \)[/tex]
#### 2. Construct the Polynomial from its Zeros
The polynomial can be written in its factored form using its zeros:
[tex]\[ f(x) = a(x - 3)(x - 2i)(x + 2i) \][/tex]
where [tex]\( a \)[/tex] is a real-valued leading coefficient.
#### 3. Simplify the Polynomial
To simplify the polynomial, first multiply the factors involving the complex zeros:
[tex]\[ (x - 2i)(x + 2i) = x^2 - (2i)^2 = x^2 - (-4) = x^2 + 4 \][/tex]
Now, the polynomial becomes:
[tex]\[ f(x) = a(x - 3)(x^2 + 4) \][/tex]
Expand this expression:
[tex]\[ f(x) = a[(x - 3)(x^2 + 4)] = a(x^3 + 4x - 3x^2 - 12) \][/tex]
[tex]\[ f(x) = a(x^3 - 3x^2 + 4x - 12) \][/tex]
#### 4. Determine the Leading Coefficient [tex]\( a \)[/tex]
We know that [tex]\( f(2) = 16 \)[/tex]. Substitute [tex]\( x = 2 \)[/tex] into the expanded polynomial and solve for [tex]\( a \)[/tex]:
[tex]\[ 16 = a(2^3 - 3(2)^2 + 4(2) - 12) \][/tex]
Calculate the values inside the parentheses:
[tex]\[ 2^3 = 8, \quad 3(2^2) = 3(4) = 12, \quad 4(2) = 8, \quad \text{and} \quad -12 = -12 \][/tex]
[tex]\[ f(2) = a(8 - 12 + 8 - 12) = 16 \][/tex]
[tex]\[ 16 = a(-8) = 16 \][/tex]
Thus:
[tex]\[ a = \frac{16}{-8} = -2 \][/tex]
#### 5. Write the Final Polynomial Function
Substitute [tex]\( a = -2 \)[/tex] back into the polynomial function:
[tex]\[ f(x) = -2(x^3 - 3x^2 + 4x - 12) \][/tex]
Distribute [tex]\( -2 \)[/tex]:
[tex]\[ f(x) = -2x^3 + 6x^2 - 8x + 24 \][/tex]
### Final Answer:
[tex]\[ f(x) = -2x^3 + 6x^2 - 8x + 24 \][/tex]
1. The polynomial has real coefficients.
2. It has zeros at 3 and [tex]\( 2i \)[/tex] where [tex]\( i \)[/tex] is the imaginary unit.
3. It satisfies the condition [tex]\( f(2) = 16 \)[/tex].
### Step-by-Step Solution:
#### 1. Determine All Zeros
First, note that since the polynomial has real coefficients and [tex]\( 2i \)[/tex] is a zero, its complex conjugate [tex]\( -2i \)[/tex] must also be a zero. Therefore, the zeros of the polynomial are:
- [tex]\( 3 \)[/tex]
- [tex]\( 2i \)[/tex]
- [tex]\( -2i \)[/tex]
#### 2. Construct the Polynomial from its Zeros
The polynomial can be written in its factored form using its zeros:
[tex]\[ f(x) = a(x - 3)(x - 2i)(x + 2i) \][/tex]
where [tex]\( a \)[/tex] is a real-valued leading coefficient.
#### 3. Simplify the Polynomial
To simplify the polynomial, first multiply the factors involving the complex zeros:
[tex]\[ (x - 2i)(x + 2i) = x^2 - (2i)^2 = x^2 - (-4) = x^2 + 4 \][/tex]
Now, the polynomial becomes:
[tex]\[ f(x) = a(x - 3)(x^2 + 4) \][/tex]
Expand this expression:
[tex]\[ f(x) = a[(x - 3)(x^2 + 4)] = a(x^3 + 4x - 3x^2 - 12) \][/tex]
[tex]\[ f(x) = a(x^3 - 3x^2 + 4x - 12) \][/tex]
#### 4. Determine the Leading Coefficient [tex]\( a \)[/tex]
We know that [tex]\( f(2) = 16 \)[/tex]. Substitute [tex]\( x = 2 \)[/tex] into the expanded polynomial and solve for [tex]\( a \)[/tex]:
[tex]\[ 16 = a(2^3 - 3(2)^2 + 4(2) - 12) \][/tex]
Calculate the values inside the parentheses:
[tex]\[ 2^3 = 8, \quad 3(2^2) = 3(4) = 12, \quad 4(2) = 8, \quad \text{and} \quad -12 = -12 \][/tex]
[tex]\[ f(2) = a(8 - 12 + 8 - 12) = 16 \][/tex]
[tex]\[ 16 = a(-8) = 16 \][/tex]
Thus:
[tex]\[ a = \frac{16}{-8} = -2 \][/tex]
#### 5. Write the Final Polynomial Function
Substitute [tex]\( a = -2 \)[/tex] back into the polynomial function:
[tex]\[ f(x) = -2(x^3 - 3x^2 + 4x - 12) \][/tex]
Distribute [tex]\( -2 \)[/tex]:
[tex]\[ f(x) = -2x^3 + 6x^2 - 8x + 24 \][/tex]
### Final Answer:
[tex]\[ f(x) = -2x^3 + 6x^2 - 8x + 24 \][/tex]
