To find a polynomial with real coefficients that has the given complex roots $3 - 2i$ and $3 + 2i$, we can use the fact that if a polynomial has real coefficients, then the complex roots must occur in conjugate pairs.

Given:

• Zeros: $3 - 2i$ and $3 + 2i$

Step-by-Step Solution

1. Write the factors associated with each root: Since $3 - 2i$ and $3 + 2i$ are roots, they correspond to the factors: $(x - (3 - 2i)) \quad \text{and} \quad (x - (3 + 2i))$

2. Multiply the factors: We want to find the product $(x - (3 - 2i))(x - (3 + 2i))$.

3. Apply the difference of squares formula: This can be simplified using the identity $(a - b)(a + b) = a^2 - b^2$.

   Let $a = x - 3$ and $b = 2i$, so: $(x - (3 - 2i))(x - (3 + 2i)) = (x - 3)^2 - (2i)^2$

4. Simplify further: $= (x - 3)^2 - 4i^2$ Since $i^2 = -1$, this becomes: $= (x - 3)^2 + 4$

5. Expand $(x - 3)^2$: $(x - 3)^2 = x^2 - 6x + 9$ So, $(x - 3)^2 + 4 = x^2 - 6x + 9 + 4 = x^2 - 6x + 13$

Final Answer

The polynomial with real coefficients that has roots $3 - 2i$ and $3 + 2i$ is: $x^2 - 6x + 13 = 0$

Would you like more details or have any questions?

Here are some related questions:

1. How do we find the polynomial if the zeros are given as complex conjugates in a different form?
2. Why must complex roots of polynomials with real coefficients come in conjugate pairs?
3. How does the difference of squares formula apply to complex conjugate pairs?
4. Can polynomials with real coefficients have purely imaginary roots?
5. How would the equation change if we were given additional real roots?

Tip: Remember, when a polynomial has real coefficients, any complex roots must appear in conjugate pairs to ensure all coefficients remain real.