# Let `P(x),deg{P(x)}=n` and `Q(x),deg{Q(x)}=m` be polynomials and suppose that ` AA a: P(a)=Q(a)`. Show that `P(x)-=Q(x)`.

Proved using mathematical induction. As for many questions about polynomials, mathematical induction is suitable here. And the parameter `n ` along which we'll move is the degree of both polynomials (actually, maximum degree of both).

To simplify the task somewhat, rewrite it this way: if `R ( x ) = 0 ` for any...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

As for many questions about polynomials, mathematical induction is suitable here. And the parameter `n ` along which we'll move is the degree of both polynomials (actually, maximum degree of both).

To simplify the task somewhat, rewrite it this way: if `R ( x ) = 0 ` for any `x , ` then the polynomial `R ` is identically zero (has all zero coefficients). Here `R ( x ) ` corresponds to `P ( x ) - Q ( x ) .`

The base of induction is the degree `n = 0 . ` A polynomial of the degree zero is a constant, `R ( x ) = c . ` If it is zero at some point, then `c = 0 , ` which is what we need.

Now let's perform the induction step from `n ` to `n + 1 . ` If a polynomial `R ( x ) ` of the degree `n + 1 ` is zero at any point, then its derivative `R' ( x ) ` is a polynomial of the degree `n ` and is also zero at any point (obvious by the definition of the derivative).

Because of this, we may apply the induction assumption to `R' ( x ) ` and conclude that `R' ( x ) ` is identically zero (has all zero coefficients). And in this case, `R ( x ) -= c = 0 , ` which is what we need.

Last Updated by eNotes Editorial on