- #1

CynicusRex

Gold Member

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## 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(x

_{1}) = Q(x

_{1}), P(x

_{2}) = Q(x

_{2}), P(x

_{3}) = Q(x

_{3}) for three different numbers x

_{1}, x

_{2}and x

_{3}, then the polynomials P(x) and Q(x) are equal.

## Homework Equations

## 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) = ax

^{2}+bx+c, Q(x) = dx

^{2}+ex+f

P(x

_{1}) = Q(x

_{1}), P(x

_{2}) = Q(x

_{2}), P(x

_{3}) = Q(x

_{3})

P(x

_{1}) - Q(x

_{1}) = 0

P(x

_{1}) - Q(x

_{1}) = P(x

_{2}) - Q(x

_{2})

(P(x

_{1}) - Q(x

_{1})) - (P(x

_{2}) - Q(x

_{2})) = 0

(P(x

_{1}) - Q(x

_{1})) - (P(x

_{2}) - Q(x

_{2})) = P(x

_{3}) - Q(x

_{3})

(P(x

_{1}) - Q(x

_{1})) - (P(x

_{2}) - Q(x

_{2})) - (P(x

_{3}) - Q(x

_{3})) = 0

P(x

_{1}) - Q(x

_{1}) - P(x

_{2}) + Q(x

_{2}) - P(x

_{3}) + Q(x

_{3}) = 0

P(x

_{1}) - P(x

_{2}) - P(x

_{3}) = Q(x

_{1}) - Q(x

_{2}) - Q(x

_{3})

$$(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(x

_{3})=Q(x

_{3})

$$P(2)=2*2^{2}-1*2+8=14\neq Q(2)=2^{2}-2*2+6=6$$

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:

_{1})=R(x

_{2})=R(x

_{3})=0;

in other words, x

_{1}, x

_{2}and x

_{3}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).