Interpolation polynomial uniqueness between two given polynomials

Question

I am trying to come up with an explanation for why too given polynomials, say $p(x) = 5x^3 - 27x^2 + 45x - 21$ and $q(x)=x^4-5x^3+8x^2-5x+3$ do not violate interpolation polynomial uniqueness.

I know these polinomial interpolate a set of values $(x_i, f(x_i)) \forall i=1..4$. (edited: changed 0 to 1, typo)

EDIT: I thought this wouldn't be needed: These polynomials interpolate this set of values: $(1,2); (2,1); (3,6); (4,47)$

I know how to proove uniqueness of interpolating polynomials, but can't see to understand how it does not get violated. How can I explain these to polynomials do not violate interpolation polynomials uniqueness?

Thanks in advance for any discussion that could lead me in the right direction.

Answer 1

The interpolation theorem tells you there is a unique cubic that meets your four points. It does not rule out higher degree polynomials also meeting your points.

For instance, for any constant $a$, $$ \frac{21+a}{24} x^4 + \frac{-45-5a}{12} x^3 + \frac{87+35a}{24} x^2 + \frac{15-25a}{12} x + a $$ meets your four points.

Your particular polynomial is $a = 3$. You get integer coefficients from $a=3 + 24k$, where $k$ is any integer. Your cubic is produced when $k = -1$. It's easy to see this is the only cubic that is produced since the quartic coefficient only has a single zero.

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The below is responsive to a previous version of the Question.

These two polynomials do not agree at all of the indicated choices for $x$. It is easy to see $p(0) = -21$ and $q(0) = 3$. (Ignore all the terms containing "$x$"s and just look at the constant term.)

Answer 2

The isomorphism only holds from $\mathbb{R}^4$ to $\mathbb{R}_3[X]$ and not $\mathbb{R}_4[X]$.