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If $f$ is a polynomial of degree $4$ such that $$f(0) = f(1) = f(2) = f(3) = 1$$ and $$f(4) = 0,$$ then determine $f(5)$.

   How would I do this? Any simple formula or theorem?
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    Similar techniques as used here: http://math.stackexchange.com/questions/2032335/constructing-a-cubic-given-four-points/. – 2017-02-12

3 Answers 3

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Define \begin{align} p(x) = f(x)-1 \end{align} then we see that $0, 1, 2, 3$ are its roots which means \begin{align} p(x) = cx(x-1)(x-2)(x-3) \end{align} where \begin{align} p(4) = f(4)-1 = -1 = c4!. \end{align} Hence $c = -1/4!$. Now finish the problem.

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    $c=\dfrac{-1}{24}$, not $\dfrac{-1}{4}$ – 2017-02-14
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    @JenkinsMa $4! = 1\times 2\times 3\times 4$. – 2017-02-14
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Hint: $$f(x)-1=a\,x\,(x-1)\,(x-2)\,(x-3)$$

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A arbitrary polynomial of degree $4$ is $p(x)=a_4x^4+a_3a^3+...a_1x+a_0.$ Put the given conditions and find the coefficients.

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    It should be helpful to insist on the fact that you generate in this way a linear system with as many equations as there are unknowns and that this system has in fact a nonzero main determinant... – 2017-02-12