Final answer:
To create a degree 3 polynomial with integer coefficients and zeros at 4i and 7/5, we must include the conjugate pair of 4i, which is -4i. Multiplying the factors for these roots and removing fractions gives the polynomial f(x) = 5x³ - 49x² + 80x - 112.
Step-by-step explanation:
To write a degree 3 polynomial with integer coefficients and with zeros at 4i and 7/5, we first need to remember that complex roots occur in conjugate pairs. Since the polynomial has integer coefficients, the third zero, which is not given, must be the conjugate of 4i, which is -4i. With this in mind, our three zeros are 4i, -4i, and 7/5.
The factors corresponding to these zeros are (x - 4i), (x + 4i), and (x - 7/5). Multiplying these factors together gives us the polynomial:
- (x - 4i)(x + 4i) = x² + 16 (since i² = -1)
- x² + 16 multiplied by (x - 7/5) will give us the required polynomial
Multiplying x² + 16 by (x - 7/5), we get:
f(x) = x³ - 73x² + 16x - 1124.
However, we want all the coefficients to be integers. To eliminate the fraction, we can multiply each term by 5, the denominator of 7/5, which will give us the polynomial with integer coefficients:
f(x) = 5x³ - 49x² + 80x - 112.