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$(-2, 8), (0, 4), (1, 2), (3, -2)$

Using the method of Newton's divided differences I found that $p(x) = 4 - 2x$ goes through these points. I have to find any other polynomial of degree 4 or less that also goes through these 4 points.

What I did was add another point and used the method of Newton's divided differences again and found a polynomial of degree 4, but it only went through the new point and not the others. It was a hellish computation too. Is there an easier way to do this?

Boot Scootin' Pootton
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2 Answers2

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Hint: let $q(x)$ be any $4^{th}$ degree polynomial which is $0$ at $x=-2,0,1,3\,$. Then $p(x)+q(x)$ will pass through those same points.

dxiv
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  • I tried this but it didn't work. I defined $q(x) = x(x+2)(x-1)(x-3)$, which is certainly zero at $x = -2, 0, 1, 3$ but the sum of this with $p(x)$ doesn't go through all of the points, unless I made a computational error. – Boot Scootin' Pootton Feb 13 '17 at 05:54
  • @BootScootin'Pootton $p(x)+q(x)=p(x)$ at all points where $q(x)=0$ so you must be making some computational error. Same goes for $p(x) + \lambda q(x)$ for any $\lambda \in \mathbb{R},$. – dxiv Feb 13 '17 at 06:04
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Hint: if $q(x)$ is a degree-four-or-less polynomial that passes through those points, then $q(x)-p(x)$ is a degree-four-or-less polynomial that has roots at $-2,0,1$ and $3$.

Arthur
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