Given the polynomials f(x)=x^3-4x^2-16x+64 and g(x)=x^2-8x+16, calculate the product f(0)*f(1)*...*f(16), if f is divisible by g.

We notice that the polynomial g(x) represents a perfect
square.

We'll re-write g =
(x-4)^2

Since f(x) is divisible by g(x), it means that we
can write the reminder theorem:

f = g*(x+a) (1), where a is
the 3rd root of f(x)

We'll factorize the first 2 terms of f
and the last 2 terms:

f =
x^2*(x-4)-16(x-4)

f = (x-4)*(x^2-16)
(2)

We'll equate (1) =
(2):

(x-4)*(x^2-16) =
g*(x+a)

We'll re-write the difference of squares from the
left:

(x-4)*(x-4)*(x+4) =
(x-4)^2*(x+a)

We'll divide by (x-4)^2 both
sideS:

x + a = x –
4

f=(x-4)^2*(x+4)

Now, we'll
calculate the product:

P =
f(0)*f(1)*...*f(16)

Since 4 is the double root of f(x), it
means that substituted in the expression of f(x), will cancel
it.

f(4) = 0

P =
f(0)*f(1)*f(2)*f(3)*f(4)...*f(16)

P =
f(0)*f(1)*f(2)*f(3)*0*..*f(16)

Since one
factor of the product is zero, the product is also zero:
f(0)*f(1)*f(2)*f(3)*f(4)...*f(16) =
0