Question 1057890
Since i is a zero, -i is also a zero, because imaginary roots always occur in conjugate pairs.


The factors of the desired polynomial are then (x - 5), (x - i), and (x + i).


The function is then:


f(x) = a(x - 5)(x - i)(x + i)


f(x) = a(x - 5)(x^2 - i^2)


f(x) = a(x - 5)(x^2 + 1)


f(x) = a(x^3 - 5x^2 + x - 5)


Then:


f(-3) = a((-3)^3 - 5(-3)^2 + (-3) - 5) = 60


a(-27 - 45 - 3 - 5) = 60


a(-80) = 60


a = -60/80


a = -3/4


The function is then:


{{{f(x) = (-3/4)x^3 + (15/4)x^2 - (3/4)x + 15/4)}}}