A fifth degree polynomial $P(x)$ with integral coefficients takes on values $0,1,2,3,4$ at $x=0,1,2,3,4$, respectively.
Which of the following is a possible value for $P(5)$?
A) $5$
B) $24$
C) $125$
D)None of the above
A fifth degree polynomial $P(x)$ with integral coefficients takes on values $0,1,2,3,4$ at $x=0,1,2,3,4$, respectively.
Which of the following is a possible value for $P(5)$?
A) $5$
B) $24$
C) $125$
D)None of the above
Any polynomial of degree at most five, fulfilling the given interpolation conditions, has the form \[ P(x) = x + a\prod_{i=0}^4 (x-i) \] for some $a \in \mathbb Z$ (we can see this if we write $P$ in the Newton basis, for example). So $P(5) = 5 + 5!a = 5 + 120a$. Now A) corresponds to $a = 0$, B) to $a = \frac{19}{120}$ and C) to $a=1$. As we want $P$ to be of fifth degree, we need $a \ne 0$, and as $\frac{19}{120} \not\in\mathbb Z$, the correct answer is C), giving $P(x) = x + \prod_{i=0}^4(x-i)$.
For example, $p\left(x\right)=x+x(x-1)(x-2)(x-3)(x-4)$.
Hint $\ $ Let $\rm\:g(x) = f(x)-x.\:$ Then
$\begin{eqnarray} &&\rm g(0)=0,\ g(1)=0,\ldots,\ g(4) = 0\\ &\iff&\rm\quad x\:|\:g,\quad\ x\!-\!1\:|\:g,\ \ldots,\ x\!-4\:|\:g \\ &\iff&\rm\quad x\,(x\!-\!1)\,\cdots\,(x\!-4)\:|\:g = f-x\\ &\iff&\rm f = x + c\, x\,(x\!-\!1)\,\cdots\,(x\!-4)\quad c\ne 0,\ deg\ c = 0\ \ by\ \ deg\ f = 5\\ &\Rightarrow&\rm f(5) = 5 + c\,5! = 5 + 120\,c\qquad excludes\ A,B,\ includes\ C \end{eqnarray}$