第一章多项式习题解答)+(-26x=2)7f(x)= g(x)(3-9+（9x9P44.1112) f(x)= g(x)(x2 +x-1)+(-5x+ 7)P44.2 1)+mx-1/x*+9x+q=余式(p+1+m)x+(q-m)=0m=qp=q2-1方法二，m-q=0x+px+q=(x2+m-1)(x+q)=[-mg-1=P同样。设2)*+m+1/x*+px*+q=余式m(p+2-m)x-(q-p+1+m)=0..m(m2+p-2)=0.m2+p=1+q,(x2=1-p+g)P44.3.1用g(x)=x+3除f(x)=2x5-5x-8x解：:. f(x)= 2(x+ 3) 30(x+3)* +175(x +3)3 495(x+ 3) + 667(x+3)327P44.3 .2)..(x3 -x2-x)=(x-1+ 2i)* +(28i)(x-1+ 2i)-(12 +8i)(x -1+2i)-(9-8i)即余式-9+8i商x-2ix-(5+2i)f(x)=x,x=1: 即P44 4.1).. f(x) =(x-1)§ + 5(x -1)* +10(x-1) +10(x-1)2 +5(x-1)+1当然也可以(x)= =[(x-1)+1)=(x-1) +5(x-1)* +..+1P44.42）结果f(x) = x* 2x2 +3 =(x+2)* - 8(x + 2) + 22(x+ 2) - 24(x+ 2)+113)f(x)= x*+2ix3 -(1+i)x2 +3x+7+i=(x+i-i)*+2i(x+i-i)3 -(1+i)(x+i-i )2-3(x+i-i)+7+i=(x+i)*-2i(x+i)*+(1+i)(x+i)-5(x+i)+7+5iP45.5(1) g(x) =(x-1)(x +2x+1)=(x-1)(x+1)2f(x)=(x+1)(3 - 3x-1):. (f(x),g(x)=x+1
第一章 多项式习题解答 P44.1 1) 1 7 26 2 ( ) ( )( ) ( ) 39 9 9 f x gx x x = − +− − + 2) 2 f x gx x x x ( ) ( )( 1) ( 5 7) = + − +− P44.2 1) 2 3 x + − + +⇒ mx x x q 1| 9 余式 2 (1 )( ) p mx qm + + +− = 0 2 1 m q p q ⎧ = ∴⎨ ⎩ = − 方法二， 设 3 2 0 ( 1)( ) 1 m q x px q x m x q mq p ⎧ − = + += + − + ⇒ ⎨ ⎩− − = 同样。 2 ) 2 42 x + + + +⇒ mx x px q 1| 余式 2 2 mp m x q p m ( 2 )( 1 ) + − − − ++ = 0 2 mm p ( 2) +− = 0. 2 2 m p qx p q 1 ,( 1 ) ∴ + =+ =− + P44.3.1 用 g() 3 x x = + 除 5 3 f () 2 5 8 xxx = −− x 解： 54 3 2 ∴ fx x x x x x ( ) 2( 3) 30( 3) 175( 3) 495( 3) 667( 3) 327 = + − + + + − + + +− P44.3 .2) 3 2 3 2 ( ) ( 1 2 ) (2 8 )( 1 2 ) x x x i ix i ∴ − − = −+ + − −+ x − + −+ − − (12 8 )( 1 2 ) (9 8 ) ix i i 即余式 −9 8 + i 商 2 x − −+ 2 (5 2 ix i) P44. 4.1). 5 0 fx x x () , 1 = = ：即 54 3 2 ∴ fx x x x x x ( ) ( 1) 5( 1) 10( 1) 10( 1) 5( 1) 1 = − + − + − + − + −+ 当然也可以 5 5 fx x x ( ) [( 1) 1] = = −+ 5 4 = − + − +⋅⋅⋅+ ( 1) 5( 1) x x 1 P44.4 2) 结果 3) 4 2 4 3 2 fx x x x x x x ( ) 2 3 ( 2) 8( 2) 22( 2) 24( 2) 11 = − += + − + + + − + + 43 2 f ( ) 2 (1 ) 3 7 x x ix i x x = + − + + ++ i 43 2 43 2 ( ) 2 ( ) (1 )( ) 3( ) 7 ( ) 2 ( ) (1 )( ) 5( ) 7 5 x i i ix i i i x i i x i i i x i ix i i x i x i i = +− + +− − + +− − +− + + = + − + + + + − + ++ P45.5 (1) 2 2 gx x x x x x ( ) ( 1)( 2 1) ( 1)( 1) = − + += − + )13)(1()( 3 xxxxf −−+= ∴( ( f x), ( )) 1 g x x = +
(2)8(x)=x*-3x2+1不可约f(x)=x4-4x+1不可约: (f(x),g(x)=1(3)(x)= x*-10x* +1=(x* +2/2x-1)(x2 -2/2x-1)g(x)=x*-4/2x*+6x* +6/2x+1, f(x)=4/2(-x* +2/2x* + x)=(x -2/2x1) .(f(x),g(x)= x2 -2/2x-1P45.6(1) /()=(x+1)°(x2 -2)g(x)=(x2 -2)(x2 +x+1) (x+1) [-(x+1)+(x2 +x+1)(x+2)=1.. (x2 -2)= -(x+1)f(x)+(x+2)g(x)(2)J(x)=(x-1)(4x3 + 2x2 -14x-y), g(x)=(x-1)(2x2 +x-4)=(x-1)f(x)=(x-1)g(x)而fi(x)= g(x)·2x-3(2x +3)gi(x)=(2x+3)·(x-1)2x1=(2x+3)(x-1)-g =(1-g3元222X-1=(x-1)f(x)+(x-1)g(x)333(3)J(x)=×4-x3 -4x2 +4x+1,g(x)=x2-x-1.. f(x)= g(x)(x2 -3)+(x-2),g(x)=(x-2)(x+1)+1: 1= -(f - g(x2 -3)(x+1)+ g= -(x+1)f(x)+(x3 + x2 - 3x-2)g(x)P45.7f(x)= g(x)1+(1+t)x2 +(2-t)x+u=r(x)11. 1-2(t2 +t+u)+(t-2)2t-2x+(1-g(x)=r(x)((+1jg)u1+t(1+t)2(1+t)由题意r(x)与g(x)的公因式应为二次所以r(x)Ig(x)t3 + 3t2 -(u + 3)t +(4 -u)(1 + t)2t? +t+3u=0. |(1+t)?1±-1否则r(x)为一次的[3 +3t? - (u +3)t +(4 -u)= 0)[(t? +t +3)u = 0
(2) 3 2 gx x x () 3 1 =− + 不可约 不可约 14)( 34 xxxf +−= ∴( ) f x gx ( ), ( ) 1 = (3) )122)(122(110)( 4 2 2 2 xxxxxxxf −−−+=+−= 4 32 32 2 gx x x x x f x x x x x x ( ) 4 2 6 6 2 1, ( ) 4 2( 2 2 ) ( 2 2 1) = − + + + = −+ + = − − 2 ∴ ( ) 2 f ( ), ( ) 2 2 1 x gx x x =− − P45.6 (1) )2()1()( 22 xxxf −+= 2 2 gx x x x ( ) ( 2)( 1) = − ++ ∵ [ ] 2 2 ( 1) ( 1) ( 1)( 2) x x xx x + − + + ++ + =1 ∴ ) 2 ( 2) ( 1) ( ) ( 2) ( x − =− + + + x f x x gx (2) )1424)(1()( ， 23 −−+−= yxxxxxf 2 gx x x x ( ) ( 1)(2 4) = − +− )()1( 1 −= xfx 1 = ( 1) ( x − g x) 而 1 1 1 ( ) ( ) 2 -3(2 3) ( ) (2 3) ( 1) fx gx x x gx x x =⋅ + = +⋅− ∴ 1 11 2 1 1 (2 3)( 1) ( )( 1) 3 3 x 1 = + −− = − −− x x g y fx g ∵ 1 22 2 1 ( 1) ( ) ( 1) ( 3 33 x − =− − + − − x f x x x gx) (3) , )( 144 234 xxxxxf ++−−= 2 gx x x () 1 = − − ∴ , 2 f x gx x x ( ) ( )( 3) ( 2) = −+ − gx x x ( ) ( 2)( 1) 1 = − ++ ∵ 2 1 ( ( 3))( 1) =− − − + + f gx x g 3 2 =− + + + − − ( 1) ( ) ( 3 2) ( x f x x x x gx) . P45.7 2 f ( ) ( )1 (1 ) (2 ) ( ) x gx tx tx u rx = ++ + − += 2 2 2 2 1 2 ( ) ( 2) 2 ( ) ( )( ) (1 ) 1 (1 ) (1 ) ( 1) t t tu t t gx rx x x u tt t t − ++ + − − = + + +− ++ + + 2 由题意 rx x () () 与g 的公因式应为二次所以rx gx ( )| ( ) ∴ ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = + ++ = + −++−+ 0 )1( 3 0 )1( )4()3(3 2 2 2 23 u t tt t ututt ∴ ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ =++ =−++−+ ⇒ −≠ 0)3( 0)4()3(3 )(1 2 23 utt ututt 否则 xrt 为一次的
解出(i)当u=0时t3 +3t2-3t+4=0(t+4)(t?-t+1)1±V3i_场1=-4或1=2.1t当u±0时，只有+1+3=0t+13(ii)P +31* -31+4 = --( + 32 - 31 + 4)t +3t?-(u+3)t+(4-u)=u=1+13[(t? +t+ 3)(t? + 2t -8)+ 6t + 24]= -2(t + 4)u=3?[u=-2(t + 4)]即(+1+3=01==1± V-112P45、8d(x)1F(x),d(x)Ig(x)表明d(x)是公因式又已知：d(x)是f(x)与g(x)的组合表明任何公因式整除d(x)所以d(x）是一个最大的公因式。P45, 9. 证明(f(x)h(x),g(x)h(x)=(f(x),g(x)h(x) (h(x)的首系=1)证: 设(f(x)h(x),g(x)h(x)=m(x)由(f(x),g(x)h(x)/f(x)h(x)(f(x),g(x)h(x)/g(x)h(x).. (f(x),g(x))h(x)|m(x): (f(x),g(x)h(x)是一个公因式。设d(x)=(f(x),g(x) =u(x)f(x)+v(x)g(x).:.d(x)h(x)=(f(x),g(x))h(x)=u(x) f(x)h(x)+v(x)g(x)h(x)而首项系数=1，又是公因式得（由P45、8），它是最大公因式，且(f(x),g(x)h(x)=(f(x)h(x),g(x)h(x)f(x)g(x)=1P45、10已知f(x),g(x)不全为0。证明(f(x),g(x)(f(x),g(x)证：设d(x)=(f(x),g(x).则d(x)±0.f(x) = f(x),g(x)=g,(x),设d(x)d(x)及d(x)=u(x)f(x)+v(x)g(x)所以 d(x)=u(x)fi(x)d(x)+v(x)g(x)d(x),消去d(x)0得1=u(x)f(x)+v(x)gi(x)P45.11 证: 设(f(x),g(x) = d(x)± 0, f(x)= f(x)d(x),g(x)= g(x)d(x). u(x)f(x)d(x)+u(x)g(x)d(x)= d(x),u(x)f(x)+u(x)g)(x)=1P45.12
解出(ⅰ)当 )1)(4(04330 23 2 时 ttttttu +−+=+−+= ∴ ¡ 3 2 ¡31 4 π ± = ± 或tt =−= e (ⅱ) 31 1 ,0 ,03 2 t t u tt −= + ≠ 只有时当 =++ )433( 31 433 )4()3(3 23 3 23 2 +−+−= + +−+ =⇒−++−+ ttt t t ttt uututt ∴ )4(2]246)82)(3[( 3 1 2 2 tttttu t +−=++−+++−= 即 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ =++ +−= 03 )4(2 2 tt tu 2 −±− 111 t = P45、8 d x( ) | ( ), ( ) | ( ) f x dx g x 表明 是公因式 d x( ) 又已知： dx f x gx () () () 是 与 的组合 表明任何公因式整除 d x( ) 所以 d x( ) 是一个最大的公因式。 P45，9. 证明( f ( ) ( ), ( ) ( )) ( ( ), ( ) ( )) xhx g xhx = f x g xhx （ 的首系 h x( ) =1） 证：设( ( ) ( ), ( ) ( )) ( ) f xhx g xhx mx = 由 ( ( ), ( )) ( ) | ( ) ( ) f x g x hx f xhx ( ( ), ( )) ( ) | ( ) ( ). f x g x hx g xhx ∴( ( ), ( )) ( ) | ( ) f x g x hx mx ∴( ( ), ( )) ( ) f x g x hx 是一个公因式。 设 d x( ) ( ( ), ( )) ( ) ( ) ( ) ( ). = =+ f x g x ux f x vx g x ∴d xhx ( ) ( ) ( ( ), ( )) ( ) ( ) ( ) ( ) ( ) ( ) ( ). = =+ f x g x hx ux f xhx vx g xhx 而首项系数=1，又是公因式得（由 P45、8），它是最大公因式，且 ( ( ), ( )) ( ) ( ( ) ( ), ( ) ( )). f x g x hx = f xhx g xhx P45、10 已知 f ( ), ( ) x g x 不全为 0。证明 () () ( , ( ( ), ( )) ( ( ), ( )) f x gx f x gx f x gx ) 1 = . 证：设 d x( ) ( ( ), ( )). = f x g x 则 d x( ) 0. ≠ 设 1 ( ) ( ), ( ) f x f x d x = 1 ( ) ( ), ( ) g x g x d x = 及 dx ux ( ) ( ) ( ) ( ) ( ). = f x vx + g x 所以 1 1 d x ux f xd x vxg xd x ( ) ( ) ( ) ( ) ( ) ( ) ( ). = + 消去 得 d x() 0 ≠ 1 1 1 () () () ( = + ux f x vxg x) P45.11 证：设 1 1 ( ( ), ( )) ( ) 0, ( ) ( ) ( ), ( ) ( ) ( ) f x gx d x f x f xd x gx g xd x =≠ = = ∴ 11 1 ux f xd x uxg xd x d x ux f x uxg x ( ) ( ) ( ) ( ) ( ) ( ) ( ), ( ) ( ) ( ) ( ) 1 += + 1 = P45.12
设uf+vg=1,uf+yh=1=uu,f?+ufyh+vguf+vugh=1: (uuf +h+vgu)f +(yu)gh=1=(f,gh)=1P45.13:(f,g)=1,固定:(f,8182)=1(f,g1-82-g,)=1P45.14(f,g)=1=uf +vg=1=(u-v)f +v(g+f)=1=(f,g+ f)=1同理(g,g+)=1由12题(Jg,f+g)=1令g=g182..gm:每个i,(f,g)=1=(ffi,g)=1,= (fiJf,g)=1=（fifzJm，8182gm）=1（注反复归纳用12题）。推广若((x),g(x)=1,则Vm,n有((x)",g(x)")=1P45,15f(x)=x°+2x*+2x+1, g(x)=x*+x*+2x*+x+1解: g(x)=f(x)(x-1)+2(x°+x+1)f(x)-(x2+x+1)(x+1)即(f(x), g(x)=x2+x+19==1+V3i6=-1-/3i622令(x2+x+1)=0得.f(x)与g(x)的公共根为,82P45.16判断有无重因式①f(x)=-5 x*+7x+2x +4x-88②(x)=x*+4x2-4x-3解①F(x)=5x4-20x3+21x2-4x+45f(x) = f(x)(x-1)-3(2x3 -5x°4x+12)1549(x2 - 4x+ 4)F(x)=(2x3 -5x2 -4x+12)(5x-+(2x35x2- 4x+12)=(x2 -4x+ 4)(2x+3)
设 2 11 1 1 1 1 uf vg u f v h uu f ufv h vgu f vu gh + = + =⇒ + + + = 1, 1 1 ∴ 11 1 1 . ( ) uu f uv h vgu f v u gh f gh + + + =⇒ = ( ) 1 ( , ) 1 P45.13 ∵ ( ,g) 1 i i f = ， 固定 1 2 :( , ) 1 i i f gg = 1 2 (, . )1 i n f ggg = P45.14 (,) 1 1 ( ) ( ) 1 (, ) 1 f g uf vg u v f v g f f g f =⇒ + =⇒ − + + =⇒ + = 同理( , gg f + =) 1. 由 12 题( , )1 fg f g + = 令 1 2 n g gg g = . ,( , ) 1 i ∴每个i fg = 1 1 ⇒ = ( ,) 1 ff g , ⇒ 1 123 ( ,) fff g = , ⇒ (, ) f f f gg g 12 1 2 " " m n =1（注反复归纳用 12 题）。 推广 若( ( ), ( )) 1, f x gx = 则∀ m，n有( (), ()) 1 m n f x gx = P45,15 f(x)=x3 +2x2 +2x+1, g(x)=x4 +x3 +2x2 +x+1 解：g(x)=f(x)(x-1)+2(x2 +x+1), f(x)=(x2 +x+1)(x+1) 即(f(x)，g(x)) = x2 +x+1. 令(x2 +x+1)=0 得 2 31 , 2 31 1 2 i −− i = +− ε = ε ∴f(x)与g(x)的公共根为 21 ε ,ε . P45.16 判断有无重因式 ① 5 432 f () 5 7 2 4 8 xx x x x x =− + + +− ② 344)( 24 xxxxf −−+= 解① 4421205)(' 4 3 2 xxxxxf +−+−= 3 2 5 ( ) '( )( 1) 3(2 5 4 12) fx f xx x x x = −− − + 2 3 2 15 49 2 '( ) (2 5 4 12)(5 ) ( 4 4) 2 2 fx x x x x x x = − −+ − + −+ )32)(44()12452( 23 2 xxxxxx ++−=+−−
故(x)有重因式(x-2)②F(x)=4x* +8x-4f(x) =(x* + 2x-1)x+(2x2 - 3x+3)f'(x)=(2x2 - 3x +3)(2x + 3)+(11x-13)) +(33+ 6x13)11(2x2 -3x +3)= (11x -13)(2x- 1111.. (f(x) f(αx)=1P45.171t=2时/(g)=x3-3x2+x-1有重因式（有重根）解. F(x)=3x2 -6x+1 3f(x)= F(x)(x-1)+(21 -6)x+(t-3)如t=3则有重因式：3重因式(x-1)"=f(x).27 ()=(2x+2)(3×-15)+(21+15如t+3.则22151=-f(x)=(x+)(x-4)42此时必须有重因式P45.18求多项式F(x)=x+px+q有重根因式的条件证f(x)=3x2 +p3f(x)=(3x + p)x+2 px+3q(p# 0)27g2(3x*+P)=(2px+3g)x- ))+(p+34p24p22p:.4p3+27q2=0p45.19 令(x)=Ax*+Bx*+1,因为(x-1)1(x),所以(x-I)I(x)即 f(x)= 4Ax3 + 2Bx =(x-1)(ax +bx+c)a=4Ab-a=0c-b=2B..4A=α=b=-2B-C=0..2A+B=0又(x-1)f(x)= f(x)=(x-1)(a'x3 +b'x +cx+ d)[a'= Ab'-a'=0.".d'=Ac'-b'=0b'= d(-d'=1-1- b'= B..A+B+1=0..A=1,B=-2x2x"f(x)=1+x+2!P46，20证n!无重因式（重根）1()=(a)-兰n!证：
故 有重因式 xf )( 3 x − )2( ② 484)(' 3 xxxf −+= 3 2 fx x x x x x ( ) ( 2 1) (2 3 3) = +− + −+ )1311()32)(332()(' 2 xxxxxf −+++−= 2 6 6 11(2 3 3) (11 13)(2 ) (33 ) 11 11 xx x x ×13 − += − − + + ∴ xfxf =1))(').(( P45.17 13)(? 有重因式（有重根） 23 时 txxxxft −+−== 解. 63)(' +−= txxxf 2 = − + − + txtxxfxf − )3()62()1)((')(3 如 则 t = 3 有重因式：3 重因式 )()1( 3 =− xfx 如 则 t ≠ .3 ) 2 15 2() 2 15 xxf 3)(22()('2 t ++−×+= 此时必须 4 15 t −= 有重因式 )4() 2 1 ()( 2 xxxf −+= P45.18 求多项式 ++= qpxxxf 有重根因式的条件 3 )( 证 += pxxf 2 3)( 2 3 ( ) (3 ) 2 3 f x x p x px = +++ q p ≠ )0( 2 2 2 2 3 3 27 (3 ) (2 3 )( ) ( ) 24 4 a q x p px q x p p p p += + − ++ 3 2 ∴4 27 p q + = 0 p45.19 令 4 2 2 f ( ) x Ax Bx x f x x f x =++ − − 1, ( 1) | ( ), ( 1) '( )) 因为 所以 即 )(1(24)(' ) 3 2 ++−=+= cbxaxxBxAxxf 4 0 2 0 a A b a cb B c ⎧ = ⎪ ⎪ − = ⎨ − = ⎪ ⎪ ⎩ − = 02 4 2 ∴ =+ ∴ = = = − BA BbaA 又 )'''')(1()()()1( 23 +++−=⇒− dxcxbxaxxfxfx ' ' ' ' '0 ' 1 a A b a c b d ⎧ = ⎪ ⎪ − = ⎨ − = ⎪ ⎪ ⎩− = 0 Bb ab Aa =−− = ∴ = '1 '' ' ∴ + BA + = 01 ∴ BA −== 2,1 P46，20 证 2 () 1 2! ! n x x fx x n =+ + +" 无重因式（重根） 证： '( ) ( ) ! n x f x fx n = −