Suppose f(x) is a polynomial of degree of 5 and with leading coefficient 2009. Supposdm further that f(1) = 1, f(2) = 3, f(3) = 5, f(4) = 7, f(5) = 9. What is the value of f(6). The answer given in book is 241091. Please work the complete solution.

Answer 1

Consider g(x) = f(x) - 2 x +1

Then g(x) = 0 for x = 1, 2,...,5, so we have:

g(x) = A (x-1)(x-2)(x-3)(x-4)(x-5)

Therefore

f(x) = A (x-1)(x-2)(x-3)(x-4)(x-5)
+ 2 x - 1

The leading coefficient is equal to A, so we have:

f(x) = 2009 (x-1)(x-2)(x-3)(x-4)(x-5)
+ 2 x - 1

Answer 2

Please explain it a bit furthur. Why did you took g(x) = f(x) - 2x + 1.

Answer 3

You have to observe pattern to get g(x) = f(x) - 2x - 1. Note that f(1) = 1 = 1+0; f(2) = 3 = 2+1, f(3) = 5 = 3+2, f(5) = 9 = 5+4. Now n+(n-1) = 2n-1. Count Iblis took g(x) = f(x) - 2x + 1 so that one has g(x) = 0 for n=1,2,...,5 then the remainder theorem can be applied.