Find a polynomial [tex]f(x)[/tex] of degree 3 with real coefficients and the following zeros.

[tex]\[
\begin{array}{l}
-3, \, 2-3i, \, 2+3i \\
f(x) = \square
\end{array}
\][/tex]



Answer :

To find a polynomial [tex]\( f(x) \)[/tex] of degree 3 with real coefficients and the given zeros, we can follow these steps:

### Step 1: List the given zeros
The zeros are:
- [tex]\(-3\)[/tex]
- [tex]\(2 - 3i\)[/tex]

### Step 2: Include the complex conjugate (since the polynomial has real coefficients)
If a polynomial has real coefficients, then the non-real zeros must come in conjugate pairs. Therefore, the conjugate of [tex]\(2 - 3i\)[/tex] is [tex]\(2 + 3i\)[/tex]. So, the zeros of the polynomial are:
- [tex]\(-3\)[/tex]
- [tex]\(2 - 3i\)[/tex]
- [tex]\(2 + 3i\)[/tex]

### Step 3: Construct the factors from the zeros
The polynomial [tex]\( f(x) \)[/tex] will be constructed from the roots:
- For zero [tex]\(-3\)[/tex], the factor is [tex]\( (x + 3) \)[/tex]
- For zero [tex]\(2 - 3i\)[/tex], the factor is [tex]\( (x - (2 - 3i)) = (x - 2 + 3i) \)[/tex]
- For zero [tex]\(2 + 3i\)[/tex], the factor is [tex]\( (x - (2 + 3i)) = (x - 2 - 3i) \)[/tex]

### Step 4: Write the polynomial as the product of these factors
[tex]\[ f(x) = (x + 3) \cdot (x - 2 + 3i) \cdot (x - 2 - 3i) \][/tex]

### Step 5: Simplify the product of the complex conjugate pair
First, simplify [tex]\( (x - 2 + 3i) \cdot (x - 2 - 3i) \)[/tex]:
[tex]\[ (x - 2 + 3i)(x - 2 - 3i) \][/tex]

This is a difference of squares:
[tex]\[ = [(x - 2)^2 - (3i)^2] \][/tex]
[tex]\[ = (x - 2)^2 - 9i^2 \][/tex]
Since [tex]\(i^2 = -1\)[/tex], we get:
[tex]\[ = (x - 2)^2 - 9(-1) \][/tex]
[tex]\[ = (x - 2)^2 + 9 \][/tex]
[tex]\[ = x^2 - 4x + 4 + 9 \][/tex]
[tex]\[ = x^2 - 4x + 13 \][/tex]

### Step 6: Multiply by the remaining factor
Now, multiply this result by the remaining factor [tex]\((x + 3)\)[/tex]:
[tex]\[ f(x) = (x + 3)(x^2 - 4x + 13) \][/tex]

Use the distributive property (FOIL method) to expand:
[tex]\[ f(x) = x(x^2 - 4x + 13) + 3(x^2 - 4x + 13) \][/tex]
[tex]\[ = x^3 - 4x^2 + 13x + 3x^2 - 12x + 39 \][/tex]
[tex]\[ = x^3 - 1x^2 + 1x + 39 \][/tex]

### Final Answer
The polynomial [tex]\( f(x) \)[/tex] with the given zeros [tex]\(-3\)[/tex] and [tex]\(2 - 3i\)[/tex] is:
[tex]\[ f(x) = x^3 - x^2 + x + 39 \][/tex]