# The polynomial f is divisible by g . Calculate p=f(0)*f(1)*...*f(2009) if f=x^3-4x^2+x+6 and g=x^2-x-2.

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f(x) = x^3-4x^2+x-6 and g(x) = x^2-x-2.

To find the product f(1) f(2)f(3)... f(2009).

Since f(x) = x^3-4x^2+x+6 is divisible g(x) = x^2-x-2,

f(x) has the same factors as that of g(x).

But x^2-x-2 = (x+1)(x-2).

Therefore x-2 factor of f(x) x^3 -4x^2+x+6.

Therefore by remaider theorem if x-2 is factor of f(x) then f(2) = 0.

Therefore the product f(1)f(2)f(3) ... f(2009) must be zero as f(2) is a factor in this product.

Therefore f(0)f(2) ... f(2009) = f(2) K = 0 as f(2) = 0.

We'll write g as a product of linear factors.

For this reason, we'll calculate it's roots applying quadratic formula:

x1 = [1 + sqrt(1 + 8)]/2

x1 = (1+3)/2

x1 = 2

x2 = (1-3)/2

x2 = -1

We'll re-write g = (x-2)(x+1)

Since f is divisible by g, it means that we can write f:

f = (x-2)(x+1)(x+a), where a is the 3rd root of f(x).

We'll compute a = 3

f = (x-2)(x+1)(x-3)

Now, we'll calculate the product:

P = f(0)*f(1)*...*f(2009)

Since 2 and 3 are the roots of f(x), it means that substituted in the expression of f(x), will cancel it.

f(2) = 0

f(3) = 0

P = f(0)*f(1)*0*0*...*f(2009)

It is enough just one factor to be zero for the product to be zero.

**P = 0**