Let a(x), b(x) belongsto R[x] two polynomials. Suppose that de[b(x)] greaterthan

Question

Let a(x), b(x) belongsto R[x] two polynomials. Suppose that de[b(x)] greaterthanorequalto 1. Show that there exist m belongsto {0, 1, ellipsis} and r_0(x), ellipsis, r_m(x) belongsto R[x] such that a(x) = r_0(x) + r_1(x)b(x) + ellipsis + r_m(x)b(x)^m and deg[r_k(x)] deg[b(x)] for all k belongsto (0, ellipsis, m). (Idea: Suppose the result is not true. Then there must be a polynomial a(x) of smallest degree such that a(x) cannot be written in the form: a(x) r_0(x) + r_1(x)b(x) + ellipsis + r_m (x) b(x)^m, with deg[r_k(x)]

Explanation / Answer

Consider a polynomial which does not satisfy the result with minimal degree be g(x). Now if degree of g(x) is less than b(x) then we can write it as r0(x) so degree of g(x) must be greater than b(x). Now devide g(x) by b(x) using Euclid's algorithm for polynomial devision. We get g(x) = b(x) q(x) + s(x) where degree of q(x) and s(x) must be less than g(x). Now degree of s(x) is less than b(x) hence it satisfies the result. In order to g(x) not satisfy the result q(x) must not satisfy the result. Hence g(x) has not minimal degree (as deg of b(x) is not 0). Hence contradicts our hypothesis hence g(x) satisfies the result.

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