Show that for any polynomial p there exists a polynomial q s

Show that for any polynomial p, there exists a polynomial q so that for all positive integers n.

Solution

If p(x)=0p(x)=0 has a root x=ax=a, then p(a)=0p(a)=0 and p(x)=p(x)p(a)p(x)=p(x)q(a).

p(x)p(a)p(x)p(a) is the sum of terms like crxrcrar=cr(xa)(xr1+axr2+a2xr3++ar1)crxrcrar=cr(xa)(xr1+axr2+a2xr3++ar1)

And since xaxa is a factor of every term, it is a factor of p(x)p(x). So every root gives us a linear factor.

Suppose p(x)=k(xa1)(xa2)…(xad)p(x)=k(xa1)(xa2)…(xad) of degree dd has the dd roots a1…ada1…ad and bb is distinct from all of these, then p(b)p(b) is a product of non-zero factors, so cannot be equal to zero.

Is this a proof without induction? Difficult to say. For example, how do we prove that if rr divides each of e1,e2…ene1,e2…en then it divides their sum?

But the division step, with p(x)=(xa)q(x)p(x)=(xa)q(x) and q(x)q(x) having strictly lower degree than p(x)p(x)leads to a descending sequence of integers (the orders of the polynomials obtained by successive divisions) - and what we need to know for that is that (i) any strictly descending sequence of non-negative integers is finite; and (ii) we can bound the length of the sequence by dd - and we can do this by observing that there are dd non-negative integers less than dd.

So it all depends on the properties of integers that we are allowed to assume - and that is because we need to say things about the degree of p(x)p(x) - an integer, and we also need to do things like indexing the coefficients.

and p1+p2........................pn=q1+q2...........................qn proved