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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
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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.

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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
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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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