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1、第一章多项式习题解答1.用g(x)除f(x)，求商q(x)与余式r(x).1）f(x)x33x2x1,g(x)3x22x13x22x1x33x2x11x7x32x21x39337x24x1337x214x739926x2991x7,r(x)26x2q(x)99.392）f(x)x42x5,g(x)x2x2x2x2x40x30x22x5x2x1x4x32x2x32x22xx3x22xx24x5x2x25x7q(x)x2x1,r(x)5x7.2.m,p,q适合什么条件时，有1）x2mx1
2、x3pxqx2
3、mx1x30x2pxqxmx3mx2xmx2(p1)xqmx2m2xm(m2p1)x(qm)当且仅当m2p10,qm时x2mx1
4、x3pxq.1本题也可用待定系数法求解.当x2mx1
5、x3pxq时，用x2mx1去除x3pxq，余式为零，比较首项系数及常数项可得其商为xq.于是有x3pxq(xq)(x2mx1)x3(mq)x2(mq1)xq.因此有m2p10,qm.2）x2mx1
6、x4px2q由带余除法可得x4px2q(x2mx1)(x2mxp1m2)m(2pm2)x(q1pm2)当且仅当r(x)m(
7、2pm2)x(q1pm2)0时x2mx1
8、x4px2q.即m(2pm2)0，即m0,或pm22,q1pm20q1p,q1.本题也可用待定系数法求解.当x2mx1
9、x4px2q时，用x2mx1去除x4px2q，余式为零，比较首项系数及常数项可得其商可设为x2axq.于是有x4px2q(x2axq)(x2mx1)x4(ma)x3(maq1)x2(amq)xq.比较系数可得ma0,maq1p,amq0.消去a可得m0,或pm22,q1q1.p,3.求g(x)除f(x)的商q(x)与余式r(x).1）f(x
10、)2x55x38x,g(x)x3;解：运用综合除法可得320508061839117327261339109327商为q(x)2x46x313x239x109，余式为r(x)327.22）f(x)x3x2x,g(x)x12i.解:运用综合除法得:12i111012i42i98i12i52i98i商为x22ix(52i),余式为98i.4.把f(x)表成xx0的方幂和，即表示成c0c1(xx0)c2(xx0)2的形式.1）f(x)x5,x01；2）f(x)x42x23,x02;3）f(x)x42ix3
11、(1i)x23x7i,x01.分析：假设f(x)为n次多项式，令f(x)c0c1(xx0)c2(xx0)2cn(xx0)nc0(xx0)[c1c2(xx0)cn(xx0)n1]c0即为xx0除f(x)所得的余式，商为q(x)c1c2(xx0)cn(xx0)n1.类似可得c1为xx0除商q(x)所得的余式，依次继续即可求得展开式的各项系数.解：1）解法一：应用综合除法得.1100000111111111111123411234513611361014114101153f(x)x5(x1)55(x1)4
12、10(x1)310(x1)25(x1)1.解法二：把x表示成(x1)1，然后用二项式展开x5[(x1)1]5(x1)55(x1)410(x1)310(x1)25(x1)12）仿上可得210203244821224112820214102421221622218f(x)1124(x2)22(x2)28(x2)3(x2)4.3）因为i12i1i37ii114ii1ii475ii01i10i5i1i1i1ii12if(x)x42ix3(1i)x23x7i(75i)5(xi)(1i)(xi)22i(xi)3
13、(xi)4.5.求f(x)与g(x)的最大公因式1）f(x)x4x33x24x1,g(x)x3x2x1解法一：利用因式分解f(x)x4x33x24x1(x1)(x33x1),4g(x)x3x2x1(x1)2(x1).因此最大公因式为x1.解法二：运用辗转相除法得q2(x)1x1x3x2x1x4x33x24x1xq1(x)24x33x21xx4x3x2x1232r1(x)2x23x18x4q3(x)2xx12x22x33221x23x1x1244x133r2(x)x044因此最大公因式为x1.2）f(
14、x)x44x31,g(x)x33x21.解：运用辗转相除法得（注意缺项系数补零）32432x4x0x0x1x1q1(x)q2(x)1x10x3x0x1393122xx43x30x2xxx33x30x2x110x22x1x33x20x133r1(x)3x2x210x210x2027x4413993x233x16256r2(x)16x11169949x21649x53916256r3(x)27256(f(x),g(x))1.3）f(x)x410x21,g(x)x442x3
