Find the polynomial function f with zeros 2/3 and 4i, and which f(3)=-525

Answer 1

If a polynomial has the zero 4i, then it automatically will also have a zero of -4i

(x + 4i)(x - 4i) = x² + 16 and (x² + 16)(x - 2/3) = x³ -2/3x² + 16x - 32/3 but this polynomial can have these same zeros if it is multiplied by a factor so:

f(x) = a(x³ - 2/3x² + 16x - 32/3)

Plugging in the point (3, -525) we get:

-525 = a(3³ - 2/3(3²) + 16(3) - 32/3)

-525 = a(27 - 6 + 48 - 32/3)

-525 = a(69 - 32/3)

-525 = a(207/3 - 32/3)

-525 = (175/3)a

a = -525(3/175) = -9

f(x) = -9(x³ - 2/3x² + 16x - 32/3)

f(x) = -9x³ + 6x² - 144x + 96