Final answer:
To find a degree 3 polynomial with integer coefficients and zeros at -4i and 9/5, one must include the conjugate of the non-real zero (4i) and multiply (x + 4i)(x - 4i)(5x - 9) to get the polynomial f(x) = 5x³ - 9x² + 80x - 144.
Step-by-step explanation:
To write a degree 3 polynomial with integer coefficients that has the zeros -4i and 9/5, we must remember that non-real zeros occur in conjugate pairs. Therefore, the polynomial will have a third zero, which is the conjugate of -4i, namely 4i. A polynomial function f(x) that satisfies these conditions can be constructed by taking the product of factors associated with each zero.
First, we express the zeros as factors in the following way:
- Zero at -4i: (x + 4i)
- Zero at 4i: (x - 4i)
- Zero at 9/5: We must multiply this by the least common multiple of its denominator to create an integer coefficient. The lowest common multiple of 5 is 5, so we have the factor of the form (5x - 9).
Next, we multiply these factors to get the polynomial:
(x + 4i)(x - 4i)(5x - 9)
The first two factors are the difference of squares and will multiply to x² + 16. We can now write:
(x² + 16)(5x - 9)
Expanding this, we get:
f(x) = 5x³ - 9x² + 80x - 144
This is the required degree 3 polynomial with the given zeros and integer coefficients.