Prove that a unique polynomial of degree n or less passes through n+1 data points.
If the polynomial is not unique, then at least two polynomials of order n or less pass
through the n+1 data points. Assume two polynomials of order n or less, Pn  x  and
Qn x  go through (n + 1) data points, x 0 , y 0 ,  x1 , y1 , ,  x n , y n 
Then define
Rn  x   Pn  x   Qn x 
Since Pn  x  and Qn  x  pass through all the (n + 1) data points,
Pn  xi   Qn  xi , i  0,  , n
Hence
Rn  xi   Pn ( xi )  Qn  xi   0, i  0,  , n
th
The n order polynomial Rn  x  has (n + 1) zeros. A polynomial of order n can have
more than n zeros (in this case n+1) only if it is identical to a zero polynomial, that is,
Rn  x   0
Hence
Pn  x   Qn  x 
Extra Notes:
How can one show that if a second order polynomial has three zeros, then it is zero
everywhere. If R2  x   a 0  a1 x  a 2 x 2 , then if it has three zeros at x1, x2, and x3, then
R1  x1   a 0  a1 x1  a 2 x12  0
R2  x 2   a 0  a1 x 2  a 2 x 22  0
R3 x3   a 0  a1 x3  a 2 x32  0
which in matrix form gives
1 x1 x12   a 0  0

  
2 
1 x 2 x 2   a1   0
1 x3 x32  a 2  0


The above set of equations has a trivial solution, that is, a0  a1  a 2  0 .
But is this the only solution?
That will be true only if the coefficient matrix is invertible.
The determinant of the coefficient matrix can be found by symbolically with forward
elimination steps of Naive Gauss elimination to give.
1 x1 x12 


det 1 x 2 x 22   x 2 x32  x 22 x3  x1 x32  x12 x3  x1 x 22  x12 x 2
1 x3 x32 


 ( x1  x 2 )( x 2  x3 )( x3  x1 )
Since
x1  x 2  x3
the determinant is non-zero. Hence the coefficient matrix is invertible. The trivial
solution a0  a1  a 2  0 is the only solution, that is, R2  x   0.