A polynomial p(x) is such that p(0) = 5, p(1) = 4, p(2) = 9, and p(3) = 20.…

We are given a polynomial p(x) with four known values:

p(0) = 5, p(1) = 4, p(2) = 9, p(3) = 20.

We want the SMALLEST degree a polynomial can have and still pass through all four points.

Key idea (finite differences): if you list the output values in order and keep subtracting neighbours, the level at which the differences become constant tells you the degree. Constant 1st differences mean degree 1, constant 2nd differences mean degree 2, and so on.

Step 1 — write the values in order of x = 0, 1, 2, 3:
5, 4, 9, 20.

Step 2 — first differences (subtract each value from the next):
4 − 5 = −1, 9 − 4 = 5, 20 − 9 = 11.

These are −1, 5, 11. They are NOT all the same, so the polynomial cannot be degree 1 (a straight line would give equal first differences).

Step 3 — second differences (subtract each first difference from the next):
5 − (−1) = 6, 11 − 5 = 6.

These are 6 and 6 — constant. Constant second differences mean a degree-2 (quadratic) polynomial is enough.

Step 4 — check with an actual quadratic. p(x) = 3x² − 4x + 5 gives:
p(0) = 5, p(1) = 3 − 4 + 5 = 4, p(2) = 12 − 8 + 5 = 9, p(3) = 27 − 12 + 5 = 20. All four match.

Since degree 1 fails but degree 2 works, the minimum degree is 2.