To find the polynomial $f(x)$ of degree $n = 4$ with the given conditions, let's proceed step by step.

Given Information:

1. Degree of Polynomial (n): 4
2. Zeros: $i$ and $2i$
3. Condition: $f(-2) = 80$

Step-by-Step Solution

Since $i$ and $2i$ are zeros and we’re dealing with a polynomial with real coefficients, the complex conjugates $-i$ and $-2i$ must also be zeros. This gives us the following zeros of the polynomial:

• $i, -i, 2i, -2i$

Forming the Polynomial

The polynomial $f(x)$ can be constructed as follows by setting up factors for each root:

$f(x) = a(x - i)(x + i)(x - 2i)(x + 2i)$

where $a$ is a constant to be determined.

Expanding the Factors

1. First, expand $(x - i)(x + i)$: $(x - i)(x + i) = x^2 + 1$

2. Next, expand $(x - 2i)(x + 2i)$: $(x - 2i)(x + 2i) = x^2 + 4$

3. Now combine these two results: $f(x) = a(x^2 + 1)(x^2 + 4)$

4. Expanding $(x^2 + 1)(x^2 + 4)$: $f(x) = a(x^4 + 4x^2 + x^2 + 4) = a(x^4 + 5x^2 + 4)$

Thus, the polynomial is: $f(x) = a(x^4 + 5x^2 + 4)$

Determining $a$ Using $f(-2) = 80$

Substitute $x = -2$ and set $f(-2) = 80$:

1. Calculate $f(-2)$ with $x = -2$: $f(-2) = a((-2)^4 + 5(-2)^2 + 4)$

2. Simplify each term: $(-2)^4 = 16, \quad (-2)^2 = 4$ $f(-2) = a(16 + 5 \cdot 4 + 4) = a(16 + 20 + 4) = a \cdot 40$

3. Set $f(-2) = 80$: $80 = a \cdot 40$ $a = \frac{80}{40} = 2$

Final Polynomial

Now we substitute $a = 2$ back into the polynomial:

$f(x) = 2(x^4 + 5x^2 + 4)$

or

$f(x) = 2x^4 + 10x^2 + 8$

Answer

The polynomial is: $f(x) = 2x^4 + 10x^2 + 8$

Would you like more details or have any questions?

Here are some related questions to further your understanding:

1. How can we verify if $f(x) = 2x^4 + 10x^2 + 8$ has $i$ and $2i$ as zeros?
2. What would happen if one of the zeros were a real number instead of complex?
3. How does the presence of complex roots impact the construction of polynomials?
4. Can this method be used to find polynomials with higher degrees?
5. How would the solution change if $f(-2)$ were given a different value?

Tip: When dealing with complex roots, always consider the conjugates to keep the polynomial coefficients real.