Write a polynomial f(x) that satisfies the given conditions.
Polynomial of lowest degree with zeros of - 3 (multiplicity 2), 2° (multiplicity 1), and with f (0) = 108.
Since the polynomial has zeros of -3 (multiplicity 2) and 2 (multiplicity 1), the factors of the polynomial are (x+3)^2 and (x-2).
f(x) = a(x+3)^2(x-2)
To find the value of 'a', we use the condition f(0) = 108:
108 = a(3)^2(-2)
108 = 9a(-2)
108 = -18a
a = -6
Therefore, the polynomial is:
f(x) = -6(x+3)^2(x-2)
f(x) = -6(x^2 + 6x + 9)(x-2)
f(x) = -6(x^3 + 6x^2 +9x - 2x^2 -12x -18)
f(x) = -6x^3 - 24x^2 -12x + 36x^2 + 144x + 108
f(x) = -6x^3 + 12x^2 + 132x + 108
So, the polynomial f(x) that satisfies the given conditions is f(x) = -6x^3 + 12x^2 + 132x + 108.