f(x) = (x - 6i)(x - (-2i))
f(x) = (x - 6i)(x + 2i)
f(x) = x² - 6xi + 2xi - 12i² (but i² = -1)
f(x) = x² - 6xi + 2xi + 12
The polynomial f(x) that satisfies the given conditions are;
f(x) = x² - 6xi + 12
We are told that the zeros of the polynomial are 6i and -2i. This means that the roots of the polynomial are 6i and -2i.
Now, we know that if we have the roots of a polynomial to be a and b, then it means the polynomial can be expressed in terms of its' factors as;
f(x) = (x - a)(x - b)
Applying the concept of factors above to our given roots and we have;
f(x) = (x - 6i)(x - (-2i)
f(x) = (x - 6i)(x + 2i)
Multiplying out gives;
f(x) = x² - 6xi + 2xi - 12i²
f(x) = x² - 6xi - 12i²
In complex numbers, i² = -1. Thus, we have;
f(x) = x² - 6xi - 12(-1)
f(x) = x² - 6xi + 12
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