Find all the zeros of f(x)=x^3+4x^2−19x+104 given that 2−3i is a zero.

Question

x=

Answer 1

By conjugate zeros theorem 2 + 3i also zero. Then f(x) is devisible by (x - 2 + 3i)(x - 2 - 3i) = x² - 4x + 13

f(x) = x³ + 4x² - 19x + 104) = (x² - 4x +13)(x + 8) = 0. So, x = 8 = 0 x = - 8

Answer: 2 - 3i, 2 + 3i and - 8 are zeros of f(x)