If $p(x)$ is a polynomial of degree $n$ such that $$p(-2)=-15,\ p(-1)=1,\ p(0)=7,\ p(1)=9,\ p(2)=13,\ p(3)=25.$$ Then smalest possible value of $n$ is

Options $(a)\; 2\;\;(b)\; 3\;\; (c)\;\; 4\;\; (d)\; 5$

Try: Tracing curve on coordinate axis, it gave one point of intersection Further $p(x)$ must be an odd degree polynomial. And slope of function is not same in each interval. So it is not linear. So it must have least degree $3$.

Can someone explain me if I am doing right? Thanks.

Otherwise please provide solution.

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    $\begingroup$ I see six points but your title only says five. $\endgroup$ – Ross Millikan Feb 12 '18 at 3:34

Construct a difference table. $$\begin{array}{rrrrr}-15&&&&\\ &16&&&\\ 1&&-10&&\\ &6&&6&\\ 7&&-4&&0\\ &2&&6&\\ 9&&2&&0\\ &4&&6&\\ 13&&8&&\\ &12&&&\\ 25&&&&\end{array}$$ Since the fourth differences are all $0$ and the third differences are not, $p(x)$ can be fitted with a third degree polynomial but not a second degree one.


There is exactly one fifth (or lower) degree polynomial passing through six points. You can find it, for example by Newton interpolation or by writing the polynomial as $ax^5+bx^4+\ldots+f$ and writing six simultaneous equations to relfect the data you have. Solve them for $a,b,c,d,e,f$. If $a \neq 0$ the polynomial has degree $5$. The fact that $p(0)=7$ gives $f=7$.

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    $\begingroup$ I'm curious if you know of a simpler method. The phrasing of the question seems to make me think it should be easier than solving so many simultaneous equations (perhaps there is a lot of cancellation that is revealed when actually writing it out, though). I otherwise agree with your answer, I just find it cumbersome. :) $\endgroup$ – Clayton Feb 12 '18 at 3:39
  • $\begingroup$ @Moo: yes, it is asking for the minimum $n$. If a cubic fits, it is the unique polynomial of fifth degree or less going through the points and the answer is $b$. Certainly one could solve for a line going through two points and see it doesn't fit a third, then the quadratic going through three and see it fails, then a cubic going through four and see it works. Whether that is easier than doing the general solution and finding $a=b=0$ I don't know. $\endgroup$ – Ross Millikan Feb 12 '18 at 4:29

You do not need to calculate any interpolation polynomial. Here, it is about the minimum degree. So the given points should have a property that makes it possible to find the degree quickly.

When you look at the points you see that the given $x$-values are consecutive integers.

So, you can make a quick check by calculating the first, second and third differences of the sequence of the $y$-values:

  • sequence: -15,1,7,9,13,25
  • first differences: 16,6,2,4,12
  • second differences: -10,-4,2,8
  • third differences: 6,6,6

As the third differences are constant the degree must be 3.


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