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f(x) is a polynomial of degree 5 with leading coeffiecient 1.f(1)=1,f(2)=4,f(3)=9,f(4)=16,f(5)=25, then f(6)=?

f(x) is a polynomial of degree 5 with leading coeffiecient 1.f(1)=1,f(2)=4,f(3)=9,f(4)=16,f(5)=25, then f(6)=?

Grade:12th pass

2 Answers

Rinkoo Gupta
askIITians Faculty 80 Points
7 years ago
From the given data it is clear that f(x)=x^2
so f(6)=(6)^2=36
Thanks & Regards
Rinkoo Gupta
AskIITians Faculty
mycroft holmes
272 Points
7 years ago
We can see that f(x) - x^2 is a polynomial of degree 5 with leading coefficient and five distinct roots namely 1,2,3,4, & 5 Hence f(x) - x^2 = (x-1)(x-2)(x-3)(x-4)(x-5) Setting x = 6, we have f(6)-6^2 = 5X4X3X2X1 = 5! = 120 so that f(6) = 120+36 = 156.

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