Let pz and qz be polynomials of degree n Show that if p and

Let p(z) and q(z) be polynomials of degree n. Show that if p and q agree at n + 1 distinct points, then p(z) = q(z) for all z.

Solution

Let p(x) be a polynomial of degree at most d, passing through d+1 distinct points, and q(x) is a different polynomial of degree d, passing through these points.

Let
r(x)=p(x)-q(x). [1.1]

r(x) is a non-zero polynomial, because p(x) and q(x) are different. Also, r(x) is a polynomial of degree at most d, because it is the difference between two polynomials of degree at most d.

According to the Fundamental Theorem of Algebra, a polynomial of degree d has exactly d roots, and, at most, d distinct roots.

At each of these d+1 distinct points where the two polynomials, p(x) and q(x), agree:
r(x)=p(x)-q(x)=0 [1.2]

Because r(x) is zero at d+1 distinct points, it has d+1 zeros (or roots). But a polynomial of degree at most d, can have at most d distinct zeros. Therefore, the two polynomials, p(x) and q(x) are not different. These polynomials are, therefore, the same polynomial. This is true because if they are not different, they can only be the same (obvious logical deduction). Associated polynomials, which differ by the multiple of a constant, agree at n points, but not at n+1.

If the polynomials are of degree less than d, then the argument is all the more powerful, because r(x) has even more \"impossible\" roots!

Whatever, of course, p(x) and q(x) will be of the same degree, because they are identical.

We can also conclude that if two polynomials, p(x) and q(x) agree at d+1 distinct points, they agree everywhere (of course, to state the obvious), because they are the same polynomial!