整除有下列性质 ③非零多项式f(x)整除其自己 Q传递性:若f(x)|g(x),g(x)|h(x),则f(x)|h(x) 若f(x)lg(x),g(x)|f(x),则f(x)=cg(x),其中0≠c∈K; Q若f(x)lg(x),i=1,2,…,m,则 f(x)|lu1(x)g1(x)+u2(x)g2(x)+…+um(x)gm(x) 其中u1(x)∈K[x] ③若f(x)|g(x),则可f(x)|g(x),其中0≠c∈K Q若g(x)|(f1(x)+22(x)且g(x)|(2f1(x)+2(x), x &(x)lf(x), 8(r)If(x) 求a,b,使x2+x+a1x3+bx+1。pê õ‘ª Ø Øke5Ÿµ ·K (Ø5Ÿ) 1 š"õ‘ª f(x) ØÙgC¶ 2 D45µe f(x) | g(x)§g(x) | h(x)§K f(x) | h(x)¶ 3 e f(x) | g(x)§g(x) | f(x)§K f(x) = cg(x)§Ù¥ 0 6= c ∈ K¶ 4 e f(x) | gi(x)§i = 1, 2, · · · , m§K f(x) | [u1(x)g1(x) + u2(x)g2(x) + · · · + um(x)gm(x)] Ù¥ ui (x) ∈ K [x]¶ 5 e f(x) | g(x)§K cf(x) | g(x)§Ù¥ 0 6= c ∈ K" ~ 1 e g(x) | (f1(x) + 2f2(x)) … g(x) | (2f1(x) + f2(x))§ K g(x) | f1(x)§g(x) | f2(x)" 2 ¦ a§b§¦ x 2 + x + a | x 3 + bx + 1"