what polynomial function with the least degree has 3i, -5 , and 5 as zeros?

Answer 1

3i, -5 , and 5 as zeros

From this, we can write the factors:

(x - 3i), (x + 3i)**, (x + 5), (x - 5)

Multiply the factors:

(x² + 9)*(x² - 25)

x⁴ - 16x² - 225)

**Note the 4th (complex) factor that was added---without this, the derived polynomial would have to have a term with i in it.

Answer 2

To answer this question, you need to know one IMPORTANT fact: complex roots (such as 3i) never come as a single root, they ALWAYS come as conjugate pairs. A conjugate is an opposite like, in this case -3i. The conjugate of "2 - 5i" is "2 + 5i". The conjugate of "-7 + i" is "-7 - i". In this case, we can write the root 3i as "0 + 3i" so the conjugate becomes "0 - 3i" our just -3i.

So if the zeros are 3i, -5 , and 5 then they MUST be (at least) 3i, -3i, -5 , and 5. If those are the zeros then we can make factors out of each one by prefacing each with "x - " like this:

(x - 3i)(x - -3i)(x - -5)(x - 5) or (x - 3i)(x + 3i)(x + 5)(x - 5) and then we just multiply them all together.

Using FOIL, we can multiply (x - 3i)(x + 3i) and get x² - 9i² (notice that the 2 middle terms drop out because they are opposites of each other) and we can simplify this by realizing that i² = -1 so it becomes x² + 9.

Then we can multiply (x + 5)(x - 5) to get x² - 25. Then lastly, we can multiply our two answers:

(x² + 9)(x² - 25) to get:

f(x) = x⁴ - 16x² - 225