# Prove: polynomial is uniquely defined by three of its values

• CynicusRex
In summary, the conversation discusses the uniqueness of a polynomial of degree not exceeding 2 when three of its values are known. Using good mathematical technique, it is shown that the problem can be simplified and generalized to polynomials of any order. The key point is that polynomial algebra often involves linear algebra, and thinking of the polynomial's coefficients as unknowns can make the problem more familiar.
CynicusRex
Gold Member

## Homework Statement

Algebra - I.M. Gelfand, Problem 164. Prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values.
This means that if P(x) and Q(x) are polynomials of degree not exceeding 2 and P(x1) = Q(x1), P(x2) = Q(x2), P(x3) = Q(x3) for three different numbers x1, x2 and x3, then the polynomials P(x) and Q(x) are equal.

## The Attempt at a Solution

The solution is simple (it's in the book) but I again approached it in a convoluted way in which I got stuck.
P(x) = ax2+bx+c, Q(x) = dx2+ex+f
P(x1) = Q(x1), P(x2) = Q(x2), P(x3) = Q(x3)

P(x1) - Q(x1) = 0
P(x1) - Q(x1) = P(x2) - Q(x2)
(P(x1) - Q(x1)) - (P(x2) - Q(x2)) = 0
(P(x1) - Q(x1)) - (P(x2) - Q(x2)) = P(x3) - Q(x3)
(P(x1) - Q(x1)) - (P(x2) - Q(x2)) - (P(x3) - Q(x3)) = 0
P(x1) - Q(x1) - P(x2) + Q(x2) - P(x3) + Q(x3) = 0
P(x1) - P(x2) - P(x3) = Q(x1) - Q(x2) - Q(x3)

$$(ax_{1}^{2}+bx_{1}+c)-(ax_{2}^{2}+bx_{2}+c)-(ax_{3}^{2}+bx_{3}+c)=(dx_{1}^{2}+ex_{1}+f)-(dx_{2}^{2}+ex_{2}+f)-(dx_{3}^{2}+ex_{3}+f)$$$$ax_{1}^{2}+bx_{1}+c-ax_{2}^{2}-bx_{2}-c-ax_{3}^{2}-bx_{3}-c=dx_{1}^{2}+ex_{1}+f-dx_{2}^{2}-ex_{2}-f-dx_{3}^{2}-ex_{3}-f$$$$a(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})+b(x_{1}-x_{2}-x_{3})-c=d(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})+e(x_{1}-x_{2}-x_{3})-f$$$$a(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})+b(x_{1}-x_{2}-x_{3})-c-d(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})-e(x_{1}-x_{2}-x_{3})+f=0$$$$(a-d)(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})+(b-e)(x_{1}-x_{2}-x_{3})-c+f=0$$Here I thought I was going to be able to prove you can only get 0 if a=d, b=e and c=f, but
$$(2-1)(4_{1}^{2}-3_{2}^{2}-2_{3}^{2})+((-1)-(-2))(4_{1}-3_{2}-2_{3})-8+6=0$$$$(1)(16-9-4)+(1)(-1)-2=0$$$$(1)(3)+(1)(-1)-2=0$$$$3-1-2=0$$
I realize I'm making a mistake by plugging in numbers, because I'm changing the given property that P(x3)=Q(x3)
$$P(2)=2*2^{2}-1*2+8=14\neq Q(2)=2^{2}-2*2+6=6$$

Again, is it possible to prove it in a similary fashion as above? I understand the given solution below, but the simplicity didn't dawn on me :(

The solution:
Consider a polynomial R(x) = P(x) - Q(x). Its degree does not exceed 2. On the other hand, we know that R(x1)=R(x2)=R(x3)=0;
in other words, x1, x2 and x3 are roots of the polynomial R(x). But a polynomial of degree not exceeding 2, as we know, cannot have more than 2 roots, unless it is equal to zero. Therefore R(x) is equal to zero and P(x) = Q(x).

There is a difference between mathematical "tricks" that are non-obvious insights that simplify a problem and good mathematical technique, which is standard and frequently repeatable, that simplifies a problem.

In this case, looking at the polynomial ##P-Q## is good technique and is something you must try to learn. It's not a one-off insight, it's a very common idea to tackle any problem where two things are equal.

PS note that you can also simply generalise the problem to polynomials of any order.

FactChecker
I think proving 'a polynomial of degree not exceeding 2 is defined uniquely by three of its values' can be done easier.

There is a key point ppl often don't realize. We think there is linear algebra, and then there is polynomial algebra which is a different thing.

Whereas in fact a lot of polynomial algebra, most of what you can do, most of what you do do, is really linear algebra!

One is in the habit of thinking of a, b, c as given knowns and x as unknown; here you just have to think of a, b, c as 'unknowns' and you have plenty of 'knowns', x1, x12, P(x1), x2, ... and looking at it that way, the problem should be something very familiar.

Last edited:

## 1. What is a polynomial?

A polynomial is a mathematical expression consisting of variables and coefficients, combined using addition, subtraction, and multiplication. It can have one or more terms and is typically written in the form of ax^n + bx^(n-1) + ... + cx + d, where a, b, c, and d are coefficients and n is a non-negative integer.

## 2. What does it mean for a polynomial to be uniquely defined?

A polynomial is uniquely defined if it can be identified or distinguished from other polynomials based on its properties, such as its degree, coefficients, and values at certain points. This means that no other polynomial can have the same properties and therefore, the polynomial in question is one-of-a-kind.

## 3. What do you mean by "three of its values"?

By "three of its values", we are referring to the values of the polynomial when the variable is substituted with three different numbers. For example, if we have the polynomial f(x) = 2x^2 + 5x + 3, then the values of f(1), f(2), and f(3) would be considered as three of its values.

## 4. How can you prove that a polynomial is uniquely defined by three of its values?

This can be proven using the Fundamental Theorem of Algebra, which states that a polynomial of degree n has exactly n complex roots. This means that if we have three distinct values of a polynomial, we can create a system of equations and solve for the coefficients, thus uniquely defining the polynomial.

## 5. Can you give an example to illustrate this concept?

Sure! Let's say we have the polynomial f(x) = 2x^2 + 5x + 3 and we know that f(1) = 5, f(2) = 16, and f(3) = 33. We can create a system of equations using these values:
f(1) = 2(1)^2 + 5(1) + 3 = 5
f(2) = 2(2)^2 + 5(2) + 3 = 16
f(3) = 2(3)^2 + 5(3) + 3 = 33
Solving this system, we get the coefficients of the polynomial as a = 2, b = 5, and c = 3, thus uniquely defining the polynomial as f(x) = 2x^2 + 5x + 3.

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