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1.

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).

Answer»

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
Discussion
2.

What do you understand by the value of polynomial at a given points?

Answer»

If p(x) is a polynomial in x, and if k is any real number, then the value obtained by replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k).
Discussion