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If it is given that f(x) is a 5th degree polynomial such that $\ f(1)=1;f(2)=3;f(3)=5;f(4)=7;f(5)=9$ what is the value of f(6)? I usually solve these kind of problems by making equations but no luck and also is there a smarter way to do this beside making 5 equations.

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  • $\begingroup$ Can be any. There's exactly one polynomial of 5th degree (maybe with zero leading coefficient) passing through six distinct points. en.wikipedia.org/wiki/… $\endgroup$ – uranix May 31 '15 at 17:23
  • $\begingroup$ If we are told the lead coefficient is $1$, we can find an answer. $\endgroup$ – André Nicolas May 31 '15 at 17:30
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HINT:

$f(x)=A(x-1)(x-2)(x-3)(x-4)(x-5)+2x-1$ where $A$ is an arbitrary constant

as the $r$th term of $1,3,5,7,9$ is $2r-1$

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Yes, there is a smarter way. We know the zeroes of $f(x)-2x+1$, they are 1,2,3,4,5. So $$f(x)-2x+1=n(x-1)(x-2)(x-3)(x-4)(x-5)$$

For some $n$. You will need to know another point or a coefficient to find the value of $f(6)$.

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