设f(x)≠0≠g(x),由带余除法有 f(x)=9(x)qn(x)+r1(x),其中 degre(x)<deg(x); g(a)=ri(a)q2(r)+r2(a), It'fdegr2()< degr1(r) rI(a)=r2(a)a3(c)+r3(x),It fdegrs(a)< degre(r); rs-2=rs-19s(a)+rs(r), It fdegrs( ) degrs-1 (a rs-1(x)=7s(x)qs+1(x)+0. 由引理知(f(x),9(x)=(f(x)-g(x)q1(x),9(x)=(r1(x),g(x)=(r1(x),r2(x) =(rs-1(x),rs(x)=cTs(x),其中c为rs(x)的首项系数的倒数.另一方面, rs(a)=Ts-2(a)-rs-1()qs(a) rs-1(x)=r-3(x)-7-2(x)qs-1(x), 所以r(x)=(1-qs-1(x)r3-2(x)-qs(x)rs-3(x),逐步代入,得 rs(r)=u(a)f()+u(a)g(a). D 下面介绍计算最大公因式的辗转相除法. 例1f(x)=x4-x3-x2+2x-1,g(x)=x3-2x+1求(f(x),g(x)和 u(a), v(a),tu( f(r)+v(r)g=(f(ar),g(a)) x)=x+1 93(a)=- r1(ar)i f(x) 6= 0 6= g(x), + f(x) = g(x)q1(x) + r1(x), [.degr1(x) < degg(x); g(x) = r1(x)q2(x) + r2(x), [.degr2(x) < degr1(x); r1(x) = r2(x)q3(x) + r3(x), [.degr3(x) < degr2(x); · · · · · · · · · · · · · · · · · · · · · rs−2 = rs−1qs(x) + rs(x), [.degrs(x) < degrs−1(x); rs−1(x) = rs(x)qs+1(x) + 0. H* (f(x), g(x)) = (f(x) − g(x)q1(x), g(x)) = (r1(x), g(x)) = (r1(x), r2(x)) = · · · = (rs−1(x), rs(x) = crs(x)), [. c ~ rs(x) !p
r!r
N,U￾ rs(x) = rs−2(x) − rs−1(x)qs(x); rs−1(x) = rs−3(x) − rs−2(x)qs−1(x), w rs(x) = (1 − qs−1(x))rs−2(x) − qs(x)rs−3(x), / d￾ rs(x) = u(x)f(x) + v(x)g(x). ✷ UCh=v33m!$1+
 1 f(x) = x 4 − x 3 − x 2 + 2x − 1, g(x) = x 3 − 2x + 1 ℄ (f(x), g(x)) 8 u(x), v(x), l u(x)f(x) + v(x)g(x) = (f(x), g(x)). x 4 −x 3 −x 2 +2x −1 x 3 −2x +1 q1(x) = x − 1 x 4 −2x 2 +x x 3 −x 2 q(x) = x + 1 −x 3 +x 2 +x −1 x 2 −2x +1 −x 3 +2x −1 x 2 −x q3(x) = −x r1(x) = x 2 −x r2 = −x +1 x 2 −x 0 2