圣才电子书www.100xuexi.com十万种考研考证电子书、题库视频学习平台第二部分课后习题第1章多项式1.用g(x)除f(x),求商q(x)与余式r(x):(1)f(x)=x3-3x2-x-1,g(x)=3x2-2x+1;(2)f(x)=x4-2x+5,g(x)=x2-x+2.解:(1)用分离系数的竖式进行计算所以17q?x?
39262
r(x)??x?
99(2)q(x)=x2+x-1,r(x)=-5x+7.2.m,p,q适合什么条件时,有1/108圣才电子书www.100xuexi.com十万种考研考证电子书、题库视频学习平台(1)x2+mx-1∣x3+px+q;(2)x2+mx+1∣x4+px2+q.解:(1)因为x3+px+q被x2+mx-1除所得的余式为(p+m2+1)x+(q-m),所以x2+mx-1∣x3+px+q的充要条件是p+m2+1=q-m=0,即p=-m2-1,q=m.(2)x2+mx+1∣x4+px2+q的充要条件是有g(x)使x4+px2+q=(x2+mx+1)g(x),比较次数及首项系数,常数项,可设g(x)=x2+ax+q.代入,展开,得?a?m?0?
?1?am?q?p?a?qm?0?
由此得x2+mx+1∣x4+px2+q的充要条件是p=2-m2,q=1或p=q+1,m=0.3.求g(x)除f(x)的商q(x)与余式r(x).(1)f(x)=2x5-5x3-8x,g(x)=x+3;(2)f(x)=x3-x2-x,g(x)=x-1+2i.解:(1)用综合除法进行计算:所以q(x)=2x4-6x3+13x2-39x+109,r(x)=-327.(2)q(x)=x2-2ix-(5+2i),r(x)=-9+8i.4.把f(x)表成x-x0的方幂和,即表成c0+c1(x-x0)+c2(x-x0)2+…的形式.(1)f(x)=x5,x0=1;2/108圣才电子书www.100xuexi.com十万种考研考证电子书、题库视频学习平台(2)f(x)=x4-2x2+3,x0=-2;(3)f(x)=x4+2ix3-(1+i)x2-3x+7+i,x0=-i.解:(1)用综合除法进行计算f?x??x4?x3?x2?x?1?x?1??1
3?2x2?3x?4?x?1??5??x?1??1??x????
2
?x3?2x2?3x?4?x?1??5?x?1??1
????????2?3x?6?x?1??10??x?1?2?5?x?1??1??x????
32
?x2?3x?6?x?1??10?x?1??5?x?1??1
??32???x?4x?1?10x?1?10x?1?5?x?1??1??????????432
??x?4??x?1??10?x?1??10?x?1??5?x?1??15432
??x?1??5?x?1??10?x?1??10?x?1??5?x?1??1
所以x5=(x-1)5+5(x-1)4+10(x-1)3+10(x-1)2+5(x-1)+1.3/108圣才电子书www.100xuexi.com十万种考研考证电子书、题库视频学习平台(2)应用综合除法所以f(x)=(x+2)4-8(x+2)3+22(x+2)2-24(x+2)+11.(3)f(x)=(x+i)4-2i(x+i)3-(1+i)(x+i)2-5(x+i)+7+5i.5.求f(x)与g(x)的最大公因式:(1)f(x)=x4+x3-3x2-4x-1,g(x)=x3+x2-x-1.(2)f(x)=x4-4x3+1,g(x)=x3-3x2+1.(3)f(x)=x4-10x2+1,g(x)?x4?42x3?6x2?42x?1解:(1)用辗转相除法进行计算4/108圣才电子书www.100xuexi.com十万种考研考证电子书、题库视频学习平台所以(f(x),g(x))=x+1.(2)(f(x),g(x))=1.(3)f?x?,g?x??x2?22x?1
??6.求u(x),v(x)使u(x)f(x)+v(x)g(x)=(f(x),g(x)):(1)f(x)=x4+2x3-x2-4x-2,g(x)=x4+x3-x2-2x-2.(2)f(x)=4x4-2x3-16x2+5x+9,g(x)=2x3-x2-5x+4.(3)f(x)=x4-x3-4x2+4x+1,g(x)=x2-x-1.解:(1)用辗转相除法进行计算.5/108