n=3; -4 and 7+4i are zeros; f(1)=260. What is f(x)

Answer 1

f(x) = (x+4)(x-7-4i)(x-7+4i) imaginary zeroes come in conjugate pairs

f(1) = 5(-6-4i)(-6+4i) = 5(36+16)=5(52)=260

f(x) = (x+4)(x^2-14x+49 +16) = (x+4)(x^2 -14x+65)

= x^3 -14x^2 +65x +4x^2 -56x +260

= x^3 -10x^2 +9x +260