The only thing I can think of is that if two polynomials are equal for all x, then they have the same degree and corresponding coeffcients are equal.
If a_0+ a_1x+ a_2x^2++ \cdot\cdot\cdot\+ a_nx^n=b_0+ b_1x+ b_2x^2+ \cdot\cdot\cdot+ b_nx^n for all x, then we must have
(a_0- b_0)+ (a_1- b_1)x+ (a_2- b_2)x^2+\cdot\cdot\cdot+ (a_n- b_n)x^n= 0 so it is sufficient to show that if
a_0+ a_1x+ a_2x^2+ \cdot\cdot\cdot+ a_nx^n= 0 for all x then a_0= a_1= a_2= \cdot\cdot\cdot= a_n= 0. That is, prove that the functions, 1, x, x^2, ..., x^n are "independent".
One way to do that is to take n different values for x, say x= 0, 1, 2, ..., n, to get n equations to solve and show that those equations are independent: x= 0 gives a_0= 0 so that's easy, x= 1 gives a_0+ a_1+ a_2+ \cdot\cdot\cdot+ a_n= 0, x= 2 gives a_0+ 2a_1+ 4a_2+ \cdot\cdot\cdot 2^n a_n= 0, etc.
More sophisticated but simpler is to note that if a_0+ a_1x+ a_2x^2+ \cdot\cdot\cdot+ a_nx^n= 0 for all x, then it is a constant so its derivative, a_1+ 2a_2x+ \cdot\cdot\cdot+ na_nx^{n-1} i also equal to 0 for all x and so its derivative if 0 for all x, etc. Setting x= 0 in the formula for the polynomial and all of its derivatives gives a_0= 0, a_1= 0, 2a_2= 0, ..., n! a_n= 0 which again say that all coefficients are 0.
