3)当0≠c∈K时,deg(cf(x)=degf(x) 命题设f(x),9(x)∈Krl,f(x)≠0,9(x)≠0,则f(x)g(x)≠0. 证明考虑首项系数 推论设∫(x)9(x)=f(x)h(x)且f(x)≠0,则g(x)=h(x) 例设∫(x),g(x)∈R[x],且f(x)2+g(x)2=0,则f(x)=9(x)=0 证明反证法假设f(x)≠0或g(x)≠0.记f(x)=anxn+an-1x2-1+…+ a1x+a9(x)=bmmm+bm-1xm-1+…+b1x+b,不妨设n≥m.则f(x)2+g(x)2 的首项系数为a2+b或a2.即f(x)2+g(x)2的首项系数不为0,与题设矛盾.口 作业:设f(x),9(x),h(x)∈R],xf2(x)+mg2(x)=h2(x),则f(x)=g(x) h(x)=0 思考题:对任意非零多项式f(x)∈K],证明degf(f(x)=(degf(x)(3)  0 6= c ∈ K U￾ deg(cf(x)) = degf(x).  T f(x), g(x) ∈ K[x], f(x) 6= 0, g(x) 6= 0,
f(x)g(x) 6= 0. $ =CZnj\
 T f(x)g(x) = f(x)h(x) L f(x) 6= 0,
g(x) = h(x).  T f(x), g(x) ∈ R[x], L f(x) 2 + g(x) 2 = 0,
f(x) = g(x) = 0. $ "!
6T f(x) 6= 0 2 g(x) 6= 0. 4 f(x) = anx n + an−1x n−1 + ... + a1x + a0 g(x) = bmx m + bm−1x m−1 + ... + b1x + b0. $T n ≥ m.
f(x) 2 + g(x) 2 Znj\g a 2 n + b 2 n 2 a 2 n . 3 f(x) 2 + g(x) 2 Znj\ g 0, {bTF
2 sT f(x), g(x), h(x) ∈ R[x], xf 2 (x) + xg2 (x) = h 2 (x),
f(x) = g(x) = h(x) = 0. ^=bPv%BnW f(x) ∈ K[x], I degf(f(x)) = (degf(x))2 . 3