Use the following proposition to prove the above proposition: Let n(x) be a poly

ID: 3078759 • Letter: U

Question

Use the following proposition to prove the above proposition:

Let n(x) be a polynomial that is not zero. For every polynomial m(x), there exist polynomials q(x) and r(x) such that

m(x) = q(x)n(x) + r(x)

and either r(x) is zero or the degree of r(x) is smaller than the degree of n(x). A root of the polynomial

p(x) = (a sub d)(x^d) + (a sub d-1)(x^(d-1)) + ... + (asub1)(x) + (a sub 0)

is a number z such that the polynomial evaluated at z is zero, that is,

p(z) =(a sub d)(z^d) + (a sub d-1)(z^(d-1)) + ... + (asub1)(z) + (a sub 0).

Assume that z is a root of p(x). Divide p(x) by (x-z). From the given proposition we know that there exists a polynomial q(x) and r(x) such that:

p(x) = (x-z)q(x) + r(x)

and

degree of r(x) < degree of (x-z) = 1 -> degree of r(x) = 0 -> r(x) is a constant number r

Therefore:

p(x) = (x-z)q(x) + r

Since z is a root of p(x): p(z) = 0

On the other hand:

(z-z)q(z) = 0*q(z) = 0

In other words:

p(z) = (z-z)q(z) + r -> 0 = 0 + r -> r = 0

So we have:

p(x) = (x-z)q(x)