I require to know the shortest way to solve it. I can solve it by taking $f(x)$ = $ax^5$ + $bx^4$ + $cx^3$ + $dx^2$ + $ex$ + $f$ and then substituting values of 1,2,3,4 and 5 in place of x and forming different equations. The equations can then be solved to find the polynomial. After finding the polynomial f(6) can be calculated. But, I want a short way to do it. The above mentioned way is very tedious and long. Please provide me with suitable answers.
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1That's insufficient information; one needs the value of a degree $5$ polynomial at six points in order to determine it. – Angina Seng May 13 '17 at 06:27
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Is $f$ specified to be monic? Or is there anything else we know about it? – Arthur May 13 '17 at 06:31
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Similar type of problem here. The methods are valid for polynomials of any degree provided you have sufficient value available. – StubbornAtom May 13 '17 at 06:37
4 Answers
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In this case any polynomial of the form $$f(x)=2x-1+b(x-1)(x-2)(x-3)(x-4)(x-5)$$ works so $f(6)$ can be anything (except $11$ which would force $f$ to have degree $1$ not $5$).
Angina Seng
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HINT:
Let $f(x)=\prod_{r=1}^6(x-r)\left[\sum_{s=1}^6\dfrac{A_s}{x-s}\right]$
lab bhattacharjee
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Consider $g(x)=f(x)-(2x-1)$. Then $g(x)$ is a polynomial of degree $5$ and it has the roots $1,2,3,4,5$. Then $f(x)=k(x-1)(x-2)(x-3)(x-4)(x-5)+(2x-1)$ for some constant $k$ which will be the leading coefficient of $g$ and hence $f$. So $f(6)=k(5!)+11$.
QED
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