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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
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