Suppose f(x) is a polynomial of degree 5 and with leading co-efficient 2009. Suppose further that f(1)=1 , f(2)=3 , f(3)=5 , f(4)=7 and f(5)=9 then find the value of f(6), f(7) and f(8).
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Let f(x) = g(x) + 2x – 1. Since f(1)=1, f(2)=3, f(3)=5, f(4)=7, f(5)=9 and 1,2,3,4,5 are all the 5 roots of g(x), then follows that g(1) = g(2) = g(3) = g(4) = g(5) = 0.
Since f(x) is 5th degree with leading coefficient 2009, the same is true for g(x).
Therefore, g(x) = 2009(x – 1)(x – 2)(x – 3)(x – 4)(x – 5).
So now, we have
f(x) = 2009(x – 1)(x – 2)(x – 3)(x – 4)(x – 5) + 2x – 1
f(6) = 2009(5)(4)(3)(2)(1) + 12 – 1 = 241091
f(7) = 2009(6)(5)(4)(3)(2) + 14 – 1 = 1446493
f(8) = 2009(7)(6)(5)(4)(3) + 14 – 1 = 5062693