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1、第一章多项式习题解答17262fxgxx()=−()()(+−x−)P44.11)399922)fxgxxx()=+()(−1)(57)+−x+232P44.21)x+−++⇒mx1
2、x9xq余式(1)()pm+++xq−=m0⎧mq=∴⎨2⎩pq=−1方法二，⎧mq−=032xpxqxm++=+−+⇒(1)(xq)⎨⎩−mq−=1p设同样。242222)x++++⇒mx1
3、xpxq余式mp(2)(+−−mxqp−+1)+=m0222∴mm(2+−=p)0.mpqx+=+1,(1=−+pq)53P44.3.1用g()xx=+3除f()2xxx=−−58x解：5432∴fx()=+−+++
4、−++2(x3)30(x3)175(x3)495(x3)667(x+−3)327P44.3.2)32∴()xx−−x32=−++−(12xii)(28)(12x−+i)−+−+−−(128)(ix12)i(98)i即余式−98+i2商x−−+2(ix52i)5.fxxx()=,=1P444.1).0：即5432∴fx()=−+−+−+−+−(x1)5(x1)10(x1)10(x1)5(x1)1+55当然也可以fxx()==−+[(x1)1]54=−+−+(1xx)5(1)⋅⋅⋅+1P44.42)结果42432fxx()=−+=+−+++−++2x3(x2)8(x2)22(x2)24(x2
5、)113)432f()xxi=+−++++2x(1)ixx37i432=+−++−−++−−+−++()xii2ixii()(1ixii)()3()xii7i432=+−++++−+++()2xiixi()(1)ixi()5()75xiiP45.522(1)gx()(1)(=−xx++=−+21)(1)(1)xxx3f(x)=(x+1)(x−3x−)1∴((fx),())gxx=+132(2)gxx()=−+3x1不可约43f(x)=x−4x+1不可约()fxgx(),()1=∴4222(3)f(x)=x−10x+1=(x+22x−1)(x−22x−)14323222gxx()=−42x
6、++6x621,()42(x+fx=−+x22xx+=−)(x221)x−∴2()f(),()xgx=−x221x−P45.62222(1)f(x)=(x+)1(x−)2gx()(=x−+2)(xx+1)22(1xxx+−+++++=)(1[)(]x1)(2x)1∵2∴(2x−=−+)(1xf)(xxg)(2++)(x)322(2)f(x)=(x−1)(4x+2x−14x−y)，gx()(1)(2=x−+xx−4)=(x−)1f(x)=(1x−)(gx)11而fxgxx()=⋅()2-3(23)x+11gx()(23)(1)=+⋅−xx121x1(23)(1)=+xxgyf−−=(−)(
7、1)xg−−1111∴331222x−=−1(xf−1)(xxxg)+(−−1)(x)∵3334322(3)f(x)=x−x−4x+4x+1,gxxx()=−−12∴fxgxx()=−()(3)(2)+x−,gx()(2)(1)1=x−+x+2∵1((3=−f−gx−))(1x++)g32=−+(1x)()(fx+xx+−32x−)(gx).P45.72f()xgx=+()1(1)++tx(2)−txurx+=()2212tt−+()t+u+(t−2)t−2gxrx()()(=+x)+x+(1−)u2221(++tt1)(1+t)(t+1)由题意rx()()与gx的公因式应为二次所以rx
8、gx()
9、()32⎧t+3t−(u+)3t+4(−u)⎫⎪2=0⎪⎪1(+t)⎪⎨⎬2⎪t+t+3⎪u=0⎪1(+t)2⎪∴⎩⎭t≠−1否则r(x)为一次的32⎧⎪t+3t−(u+)3t+4(−u)=0⎫⎪⇒⎨⎬2⎪⎩(t+t+)3u=0⎪⎭∴322解出(ⅰ)当u=0时t+3t−3t+4=(0t+4)(t−t+)1π1±¡3±¡t=−4或t==e3∴221t当u≠0时,只有t+t+3=,0=−(ⅱ)t+13t3+3t2−3t+4t3232t+3t−(u+)3t+4(−u)⇒u==−(t+3t−3t+)4t+13122u=−[(t+t+3)(t+2t−)8+6t+24]=−(2t+)4∴3
10、⎧u=−(2t+)4⎫⎨2⎬即⎩t+t+3=0⎭−1±−11t=2P45、8dx()
11、(),()
12、()fxdxgx表明dx()是公因式又已知：dxfxgx()()是与()的组合表明任何公因式整除dx()所以dx()是一个最大的公因式。P45，9.证明(f()(),()())((),()())xhxgxhx=fxgxhx（hx()的首系=1）证：设(()(),()())fxhxgxhx=mx()由((),())()
13、()()fxgxh
