f(x)=x^3+2x^2−14x+272 given that 3−5i is a zero.

Question

x=

Answer 1

By conjugate zeros theorem 3 + 5i also zero. Then f(x) divisible by (x - 3 - 5i)(x - 3 + 5i) = x² - 6x + 34 and

f(x) = (x² - 6x + 34)(x + 8) = 0; x + 8 = 0; x = - 8

Answer: 3 - 5i, 3 + 5i and - 8 are zeros of f(x)