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.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

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