Given that x=-1+4i is a zero of f (x)= x^3+x^2+15x-17 find all the zeroes of f

Answer:All the zeroes of f(x) are x = 1, x = -1 + 4i and x = -1 - 4i.Step-by-step explanation:Given that f(x) = x³ + x² + 15x - 17 Now, we have to find all the zeroes of the function. Given that x = - 1 + 4i is a zero of the function. So, x = - 1 - 4i must be another zero of the function. Therefore, (x + 1 - 4i)(x + 1 + 4i) will be factor of the function. Hence,  (x + 1 - 4i)(x + 1 + 4i) = x² + 2x + (1 - 4i)(1 + 4i)  = x² + 2x + [1² - (4i)²] = x² + 2x + 17 Assume that (x + a) is another factor of f(x). Therefore, we can write f(x) = x³ + x² + 15x - 17 = (x + a)(x² + 2x + 17) ⇒ x³ + x² + 15x - 17 = x³ + (a + 2)x² + (2a + 17)x + 17a Hence, comparing the coefficients we can write  a + 2 = 1  ⇒ a = -1  Therefore, f(x) =x³ + x² + 15x - 17 = (x - 1)(x² + 2x + 17) So, all the zeroes of f(x) are x = 1, x = -1 + 4i and x = -1 - 4i (Answer)