《高等代数与解析几何》课程教学资源（习题解答）第十章 一元多项式与整数的因式分解

习题解答 第十章一元多项式与整数的因式分解 习题10-1 1.计算(x2+ax-b)(x2-1)+(x2-ax+b)(x2+1) 解:2x4-2ax+ 2.计算多项式x3+2x2+3x-1与3x2+2x+4的乘积 解:3x5+8x4+17x3+11x2+10x-4 f(x)=3x2-5x+3 9(x)=ax(x-1)+b(x+2)(x-1)+cr(x+2) 试确定a,b,c,使f(x)=g(x) 解:取x 4.设∫(x),g(x)和h(x)都是实系数多项式证明:如果 f(a)=rg(a)+rh(a), 那么 f(x)=9(x)=h(x) 证明:如f(x)≠0,则左式的次数为偶数,而右式的次数为奇数,矛盾,故f(x)=0.从而 又,g(x),h(x)皆为实系数多项式,从而g2(x),h2(x)的首项系数都是非负数而这两个数之和为零,故 g(x),h(x)的首项系数都是零,从而g(x)=h(x)=0 习题102 1.用g(x)除f(x)求商q(x)与余式r(x) (1)f(x)=x4+4x2-x+6,g(x)=x2 (2)f(x)=x3+3x2-x-1,g(x)=3x2-2x+1 解:(1)q(x)=x2-x+4,r(x)=-4x+2 ()=号(3 (x-2) 2.m,P,q适合什么条件时,有 (2) 解:(1)p=1-m2, 或

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 % &￾'() % & 10–1 1. op (x 2 + ax − b)(x 2 − 1) + (x 2 − ax + b)(x 2 + 1). ): 2x 4 − 2ax + 2b. 2. opI~ x 3 + 2x 2 + 3x − 1  3x 2 + 2x + 4  . ): 3x 5 + 8x 4 + 17x 3 + 11x 2 + 10x − 4. 3. f(x) = 3x 2 − 5x + 3, g(x) = ax(x − 1) + b(x + 2)(x − 1) + cx(x + 2), <fX a, b, c, N f(x) = g(x). ): Q x = −2, O a = 25 6 ; Q x = 0, O b = − 3 2 , Q x = 1, O c = 1 3 . 4. f(x), g(x) u h(x) 2!syI~, ^_: 45 f 2 (x) = xg2 (x) + xh2 (x), [# f(x) = g(x) = h(x) = 0. '(: 4 f(x) 6= 0, y.y, Jy.y, , 3 f(x) = 0. IJ g 2 (x) + h 2 (x) = 0. C, g(x), h(x) .syI~, IJ g 2 (x), h2 (x) !~sy2!zzy, Jk@vytu.{, 3 g(x), h(x) !~sy2!{, IJ g(x) = h(x) = 0. % & 10–2 1. X g(x) " f(x), L# q(x) m r(x): (1) f(x) = x 4 + 4x 2 − x + 6, g(x) = x 2 + x + 1; (2) f(x) = x 3 + 3x 2 − x − 1, g(x) = 3x 2 − 2x + 1. ): (1) q(x) = x 2 − x + 4, r(x) = −4x + 2. (2) q(x) = 1 9 (3x + 11), r(x) = 10 9 (x − 2). 2. m, p, q $%"#FGS, $ (1) x 2 + mx + 1 | x 3 + px + q; (2) x 2 + mx + 1 | x 4 + px2 + q. ): (1) p = 1 − m2 , q = −m. (2) ( m = 0 p = 1 + q k ( p = −m2 + 2 q = 1 · 1 ·

3.用综合除法求商q(x)及余式r(x) (1)f(x)=x4-2x3+4x2-6x+8,g(x)=x-2; (2)f(x)=2x5-5x3-8x,g(x)=x+2 解:(1)q(x)=x3+4x+2,r(x)=12 (2)q(x) 4.用综合除法表f(x)为x-x0的方幂 (1)f(x)=x4-2x3+3x2-2x+1,xo=2; (2)f(x)=x4-2x2+3,xo=-2 (3)f(x)=x4+2in3-(1+i)x2-3x+1 解:(1)f(x)=(x-2)4+6(x-2)3+15(x-2)2+18(x-2)+9 (2)f(x)=(x+2)4-8(x+2)3+22(x+2)2-24(x+2)+1 (3)f(x)=(x+i)4-2i(x+i)3-(1+i)(x+i)2-5(x+i)+(1+2i) 5.记(x)°=1,(x)=x(x-1)(x-2)…(x-k+1),(k>1).试将f(x)表为 的形式 (1)f(x)=x4-2x3+x2-1 (2)f(x)=x5 解:(1)1 2 3112 因此f(x)=-1+2({x)2+4(x)3+(x)4 (2)f(x)={x)+15(x)2+25{x)3+10(x)4+(x) 6.k是正整数,证明:x|f(x)当且仅当x|f(x) 证明:设f(x)的常数项为a,则(x)的常数项为a因此x|f(x)→ak=0→a=0←→ I f(r) 7.设a,b为两个不相等的常数,证明:多项式f(x)被(x-a)(x-b)除所得余式为 f(a)-f(b) af(b)-bf(a 证明:设f(x)=(x-a)(x-b)q(x)+Ax+B,则 f(a)=aA+B, f(b=bA+B 由此得 f(a)-f(b) f(b)-bf(a A 因此结论成立 8.设f1(x),f2(x),91(x),9(x)都是数域K上的多项式,其中f1(x)≠0. 证明:如果g1(x)g2(x)|f1(x)f2(x),f1(x)|g1(x),则g2(x)|f2(x)

3. X&%"qL# q(x) .m r(x): (1) f(x) = x 4 − 2x 3 + 4x 2 − 6x + 8, g(x) = x − 2; (2) f(x) = 2x 5 − 5x 3 − 8x, g(x) = x + 2. ): (1) q(x) = x 3 + 4x + 2, r(x) = 12. (2) q(x) = 2x 4 − 4x 3 + 3x 2 − 6x + 4, r(x) = −8. 4. X&%"qr f(x) . x − x0 >': (1) f(x) = x 4 − 2x 3 + 3x 2 − 2x + 1, x0 = 2; (2) f(x) = x 4 − 2x 2 + 3, x0 = −2; (3) f(x) = x 4 + 2ix 3 − (1 + i)x 2 − 3x + 1 − 2i, x0 = −i. ): (1) f(x) = (x − 2)4 + 6(x − 2)3 + 15(x − 2)2 + 18(x − 2) + 9. (2) 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) + (1 + 2i). 5. G hxi 0 = 1, hxi k = x(x − 1)(x − 2)· · ·(x − k + 1), (k > 1). <( f(x) r. c0 + c1hxi + c2hxi 2 + · · · p: (1) f(x) = x 4 − 2x 3 + x 2 − 1; (2) f(x) = x 5 . ): (1) 1 1 −2 1 0 −1 1 −1 0 2 1 −1 0 0 2 2 3 1 1 2 3 1 4 l f(x) = −1 + 2hxi 2 + 4hxi 3 + hxi 4 . (2) f(x) = hxi + 15hxi 2 + 25hxi 3 + 10hxi 4 + hxi 5 . 6. k !W y, ^_: x | f k (x) PFaP x | f(x); '(: f(x) `y~. a,  f k (x) `y~. a k . l x | f k (x) ⇐⇒ a k = 0 ⇐⇒ a = 0 ⇐⇒ x | f(x). 7. a, b .@v``y, ^_: I~ f(x) ) (x − a)(x − b) "Om. f(a) − f(b) a − b x + af(b) − bf(a) a − b . '(: f(x) = (x − a)(x − b)q(x) + Ax + B,  f(a) = aA + B, f(b) = bA + B, lO A = f(a) − f(b) a − b , B = af(b) − bf(a) a − b . l89:;. 8. f1(x), f2(x), g1(x), g2(x) 2!y
K I~, -* f1(x) 6= 0. ^_: 45 g1(x)g2(x) | f1(x)f2(x), f1(x) | g1(x),  g2(x) | f2(x). · 2 ·

证明:设f1(x)f2(x)=91(x)g2(x)q1(x),g1(x)=f1(x)q2(x).则f1(x)f2(x)=f1(x)q(x)92(x)q1(x) 由于f1(x)≠0,两得f2(x)=92(x)q2(x)q1(x),即g2(x)|f2(x) 9.证明:x4-1|x2-1当且仅当d|n. 证明:(→)若n=dq,则 xn-1=(x4-1)(x 因此x4-1|xn-1. (←)设n=d+r,0≤r<d.由上证,x-1≡0(modx2-1).即 1) 1≡x-1(1 而x-1|x-1台r=0.因此x-1|xn-1兮r=0台d|n 10-3 1.求最大公因式(f(x),9(x) (1)f(x)=x4+x3-3x2-4x-1,g(x)=x3+x2-x-1 1,g(x) (3)f(x)=x4-x3-4x2+4x+1,g(x)=x2-x-1 解:(1)x (2)1 2.求u(x),v(x),使u(x)f(x)+v(x)g(x)=(f(x),9(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)=2x4+3x3-3x2-5x+2,g(x)=2x3+x2-x-1 解:(1)u(x)=-x-1,v(x)=x+2,d(x)=x2-2 (2)u()=-(x-1),v(x)=(2x2-2x-3),dx)=x-1 )u()=-(2x2+3m,()=(2x3+52-6,d(x)=1 3.证明:如果d(x)|∫(x),d(x)lg(x),且d(x)为f(x)与9(x)相一个组合,那么d(x)是f(x)与g(x) 相一个最大公因式 证明:设d(x)=u(x)f(x)+u(x)9(x),则对任意相h(x)∈K[x,如h(x)|f(x),h(x)|g(x), h(a)l d(a) 又,d(x)为f(x)与g(x)相一个公因式,故d(x)是f(x)与g(x)相一个最大公因 证明:如果h(x)为首一多项式,则 (f(x)h(x),9(x)h(x)=(f(x),g(x)h(x) 证明:设d(x)=(f(x),9(x)≠0,则存法u(x),v(x)使 d(e=ur)f()+u(a)g(r

'(: f1(x)f2(x) = g1(x)g2(x)q1(x), g1(x) = f1(x)q2(x).  f1(x)f2(x) = f1(x)q2(x)g2(x)q1(x),  f1(x) 6= 0, @O f2(x) = g2(x)q2(x)q1(x), K g2(x) | f2(x). ∗9. ^_: x d − 1 | x n − 1 PFaP d | n. '(: (⇒)  n = dq,  x n − 1 = (x d − 1)(x d(q−1) + x d(q−2) + · · · + x d + 1). l x d − 1 | x n − 1. (⇐) n = dq + r, 0 6 r < d. ^, x dq − 1 ≡ 0 (mod x d − 1). K x dq ≡ 1 (mod x d − 1), x n ≡ x dq+r ≡ x dq · x r ≡ x r (mod x d − 1), x n − 1 ≡ x r − 1 (mod x d − 1). J x d − 1 | x r − 1 ⇔ r = 0, l x d − 1 | x n − 1 ⇔ r = 0 ⇔ d | n. % & 10–3 1. L^| (f(x), g(x)): (1) f(x) = x 4 + x 3 − 3x 2 − 4x − 1, g(x) = x 3 + x 2 − x − 1; (2) f(x) = x 5 + x 4 − x 3 − 2x − 1, g(x) = 3x 4 + 2x 3 + x 2 − 2; (3) f(x) = x 4 − x 3 − 4x 2 + 4x + 1, g(x) = x 2 − x − 1. ): (1) x + 1. (2) 1. (3) 1. 2. L u(x), v(x), N u(x)f(x) + v(x)g(x) = (f(x), g(x)): (1) f(x) = x 4 + 2x 3 − x 2 − 4x − 2, g(x) = x 4 + x 3 − x 2 − 2x − 2; (2) f(x) = 4x 4 − 2x 3 − 16x 2 + 5x + 9, g(x) = 2x 3 − x 2 − 5x + 4; (3) f(x) = 2x 4 + 3x 3 − 3x 2 − 5x + 2, g(x) = 2x 3 + x 2 − x − 1. ): (1) u(x) = −x − 1, v(x) = x + 2, d(x) = x 2 − 2. (2) u(x) = − 1 3 (x − 1), v(x) = 1 3 (2x 2 − 2x − 3), d(x) = x − 1. (3) u(x) = − 1 6 (2x 2 + 3x), v(x) = 1 6 (2x 3 + 5x 2 − 6), d(x) = 1. 3. ^_: 45 d(x) | f(x), d(x) | g(x), F d(x) . f(x)  g(x) fv]%, [# d(x) ! f(x)  g(x) fv^|. '(: d(x) = u(x)f(x) + v(x)g(x), 45 h(x) ∈ K[x], 4 h(x) | f(x), h(x) | g(x),  h(x) | d(x). C, d(x) . f(x)  g(x) fv, 3 d(x) ! f(x)  g(x) fv^|. 4. ^_: 45 h(x) .!fI~,  (f(x)h(x), g(x)h(x)) = (f(x), g(x))h(x). '(: d(x) = (f(x), g(x)) 6= 0, Dq u(x), v(x) N d(x) = u(x)f(x) + v(x)g(x). · 3 ·

所以 d(x)h(x)=u(x)f(x)h(x)+v(x)g(x)h(x) 又因d(x)h(x)|f(x)h(x),d(x)h(x)|g(x)h(x),所以d(x)h(x)是∫(x)h(x)与g(x)h(x)的一个最大公因式 又因d(x),h(x)都是首一多项式故d(x)h(x)也是首一多项式,从而 (f(x)h(x),g(x)h(x)=d(x)h(x)=(f(x),9(x)h(x) 又如d(x)=0,则f(x)=9(x)=0.原等式仍然成立 5.证明:如果f(x),g(x)不全为零,则 f(r) ga (f(x,(x)(f(x),g(z)/1. 证明:因f(x),9(x)不全为零,故(f(x),9(x)≠0.所以 f(x),9(x) (uiah.ole)((e), g(2). ig). ao)U(a) g(e) f(a) g(r) (f(x),9(x)(f(x),g(x) f(x),9(x) (由习题4)两边消去(f(x),g(x),得 g(a) (f(x),g(x)’(f(x),9(x) 6.证明:如果∫(x),9(x)不全为零,且 u(a)f(a)+v(a)g(a)=(f(r),g() 则(u(x),v(x)=1. 证明:因f(x),9(x)不全为零,故(f(x),9(x)≠0.因此 u(a)(x)+()(2) (f(x),9(m), (a(x),v(x)=1 7.证明:如果(f(x),9(x)=1,(f(x),h(x)=1,那么 (x),9(x)h(x) 证明:存在u(x),U(x),s(x),t(x),使 u(x)f(x)+v(x)9(x)=1 s(a)f(a)+t(a)h(a)=l ∫(x)(u(x)s(x)f(x)+u(x)t(x)h(x)+(x)v(x)9(x))+v(x)t(x)9(x)h(x)=1, (f(x),g(x)h(x)=1. 8.设f1(x),…,fn(x),91(x),…,9n(x)都是多项式,且(f1(x)9(x)=1(i=1,……,m;j 1,…,n),证明 (f1(x)f2(x)…fm(x),91(x)g2(x)…gn(x)=1 证明:由(f(x),9(x)=1,可得(f(x),g1(x)g2(x))=1,……,(f(x),g1(x)g2(x)…gn(x))=1.从而 (f1(x)f2(x),g1(x)…9n(x))=1,(f1(x)f2(x)f3(x),g1(x)…9n(x)=1,…, (f1(x)f2(x)…fm(x),91(x)…9n(x)=1. 9.证明:如果(f(x),g(x)=1,那么(f(x)+9(x),f(x)9(x)=1

 d(x)h(x) = u(x)f(x)h(x) + v(x)g(x)h(x). C d(x)h(x) | f(x)h(x), d(x)h(x) | g(x)h(x),  d(x)h(x) ! f(x)h(x)  g(x)h(x) fv^|. C d(x), h(x) 2!!fI~, 3 d(x)h(x) ;!!fI~, IJ (f(x)h(x), g(x)h(x)) = d(x)h(x) = (f(x), g(x))h(x). C4 d(x) = 0,  f(x) = g(x) = 0, r*:;. 5. ^_: 45 f(x), g(x) `y.{,  µ f(x) (f(x), g(x)), g(x) (f(x), g(x))¶ = 1. '(:  f(x), g(x) `y.{, 3 (f(x), g(x)) 6= 0.  (f(x), g(x)) = µ f(x) (f(x), g(x)) (f(x), g(x)), g(x) (f(x), g(x)) (f(x), g(x))¶ = µ f(x) (f(x), g(x)) , g(x) (f(x), g(x)) ¶ (f(x), g(x)) (NO 4) @R+, (f(x), g(x)), O µ f(x) (f(x), g(x)) , g(x) (f(x), g(x)) ¶ = 1. 6. ^_: 45 f(x), g(x) `y.{, F u(x)f(x) + v(x)g(x) = (f(x), g(x)),  (u(x), v(x)) = 1. '(:  f(x), g(x) `y.{, 3 (f(x), g(x)) 6= 0, l u(x) f(x) (f(x), g(x)) + v(x) g(x) (f(x), g(x)) = 1, (u(x), v(x)) = 1. 7. ^_: 45 (f(x), g(x)) = 1, (f(x), h(x)) = 1, [# (f(x), g(x)h(x)) = 1. '(: Dq u(x), v(x), s(x), t(x), N u(x)f(x) + v(x)g(x) = 1, s(x)f(x) + t(x)h(x) = 1,  f(x)(u(x)s(x)f(x) + u(x)t(x)h(x) + s(x)v(x)g(x)) + v(x)t(x)g(x)h(x) = 1, (f(x), g(x)h(x)) = 1. 8. f1(x), · · · , fm(x), g1(x), · · · , gn(x) 2!I~, F (fi(x), gj (x)) = 1 (i = 1, · · · , m; j = 1, · · · , n), ^_: (f1(x)f2(x)· · · fm(x), g1(x)g2(x)· · · gn(x)) = 1. '(:  (fi(x), gj (x)) = 1, @O (fi(x), g1(x)g2(x)) = 1, . . . , (fi(x), g1(x)g2(x)· · · gn(x)) = 1. IJ (f1(x)f2(x), g1(x)· · · gn(x)) = 1, (f1(x)f2(x)f3(x), g1(x)· · · gn(x)) = 1, . . . , (f1(x)f2(x)· · · fm(x), g1(x)· · · gn(x)) = 1. 9. ^_: 45 (f(x), g(x)) = 1, [# (f(x) + g(x), f(x)g(x)) = 1. · 4 ·

证明:由于(f(x),g(x)=1,所以 (f(x)+g(x),g(x)=(f(x),9(x))=1 (f(x)+g(x),f(x))=(g(x),f(x)=1 因此 (f(x)+9(x),f(x)g(x)=1 10.设f1(x)=af(x)+bg(x),91(x)=cf(x)+dg(x),且ad-be≠0,证明 (f(x),g(x)=(f1(x),91(x) 证明:由题设可得(f(x),g(x)(f1(x),91(x).又 f(a) ad-bc i(z)--6 f1(x)+ 91(a) 所以 (f1(x),g1(x)|(f(x),g(x) 又因(f1(x),91(x)与(f(x),9(x)的首项系数相同,故(f(x),g(x)=(f1(x),g1(x) 11.证明:如果f(x)与g(x)互素,那么f(xm)与g(m)也互素 证明:由题设,存在多项式u(x),v(x)使 u(x)f(x)+v(x)9(x)=1 所以 u(zm)f(am)+v(am)g(rm)=l 故(f(xm),g(xm) 12.证明:对任意的正整数m,都有 (f(x),g(x)=(f(x),g"(x)) 证明:设(f(x),9(x)=d(x),f(x)=d(x)f1(x),9(x)=d(x)g1(x),则(f1(x),g1(x) 由习题8可得 (f(x),g1(x)=1 于是 (尸"(x),g"(x))=(d(x)f(x),d"(x)9i(x) d(x)((x),(x)=d(x) *13.试求xm-1与xn-1的最大公因式 解:令d=(m,n),则根据习题10-2.9,x4-1 1,x4-1|xn-1 设h(x)是mm-1与x-1的公因式,则有 m-1≡0(modh(x),xn-1≡0(modh(x)→mm≡1(modh(x),xn≡1(modh(x) 由于d=(m,m),因此存在u,v∈Z使得d=um+m 1 (mod h(a))=> 1≡0(modh(x)

'(:  (f(x), g(x)) = 1,  (f(x) + g(x), g(x)) = (f(x), g(x)) = 1, (f(x) + g(x), f(x)) = (g(x), f(x)) = 1, l (f(x) + g(x), f(x)g(x)) = 1. 10. f1(x) = af(x) + bg(x), g1(x) = cf(x) + dg(x), F ad − bc 6= 0, ^_: (f(x), g(x)) = (f1(x), g1(x)). '(: O @O (f(x), g(x)) | (f1(x), g1(x)). C f(x) = d ad − bc f1(x) − b ad − bc g1(x), g(x) = −c ad − bc f1(x) + a ad − bc g1(x),  (f1(x), g1(x)) | (f(x), g(x)). C (f1(x), g1(x))  (f(x), g(x)) !~sya, 3 (f(x), g(x)) = (f1(x), g1(x)). 11. ^_: 45 f(x)  g(x)
, [# f(x m)  g(x m) ;
. '(: O , DqI~ u(x), v(x) N u(x)f(x) + v(x)g(x) = 1.  u(x m)f(x m) + v(x m)g(x m) = 1. 3 (f(x m), g(x m)) = 1. 12. ^_ : 45W y n, 2$ (f(x), g(x))n = (f n (x), gn (x)). '(: (f(x), g(x)) = d(x), f(x) = d(x)f1(x), g(x) = d(x)g1(x),  (f1(x), g1(x)) = 1. NO 8 @O (f n 1 (x), gn 1 (x)) = 1. ! (f n (x), gn (x)) = (d n (x)f n 1 (x), dn (x)g n 1 (x)) = d n (x)(f n 1 (x), gn 1 (x)) = d n (x) = (f(x), g(x))n . ∗13. <L x m − 1  x n − 1 ^|. ): S d = (m, n), x-NO 10–2.9, x d − 1 | x m − 1, x d − 1 | x n − 1. h(x) ! x m − 1  x n − 1 , $ x m − 1 ≡ 0 (mod h(x)), xn − 1 ≡ 0 (mod h(x)) =⇒ x m ≡ 1 (mod h(x)), xn ≡ 1 (mod h(x)).  d = (m, n), lDq u, v ∈ Z NO d = um + vn. x d = x um+vn ≡ 1 (mod h(x)) =⇒ x d − 1 ≡ 0 (mod h(x)). · 5 ·

又设d=ms-nt,s,t≥0,则d+nt=ms.于是 rms-1=rd+nr-1=(ard-1)xr+xnr-1 若f(x)∈K[]满足f(x)|xm-1,f(x)|xn-1,则(f(x),x)=1,且f(x)|xm-1,f(x)|xm-1,于是 f(x)|(x4-1)xm.由f(x)与x互素可得f(x)|x2-1.因此(xm-1,xm-1)=x4-1,其中d=(m,n 1.证明只要(1,=0(的次数都大于零,就可以适当选择适合等式 u(x)f(x)+v(x)g(x)=(f(x),9(x) 的u(x)与v(x),使 deg u(a)f(r) f(r),g(a)) 习题10-4 1.设(f(x),m(x)=1,证明:对任何的多项式g(x),都存在多项式h(x),使 h(a)f(a)=g(r)(mod m(r)) 证明:由假设,存在u(x),v(x)∈K],使 u(a)f(a)+v(a)m(a)=1 g(a)u(a)f(r)+g(a)v(e)m(a)=g(a). 于是 g(r)u(a)f(=g(r(mod m())

C d = ms − nt, s, t > 0,  d + nt = ms. ! x ms − 1 = x d+nr − 1 = (x d − 1)x nr + x nr − 1.  f(x) ∈ K[x]  f(x) | x m − 1, f(x) | x n − 1,  (f(x), x) = 1, F f(x) | x ms − 1, f(x) | x nt − 1, ! f(x) | (x d − 1)x nr .  f(x)  x
@O f(x) | x d − 1. l (x m − 1, xn − 1) = x d − 1, -* d = (m, n). ∗14. ^_: lE f(x) (f(x), g(x)) , g(x) (f(x), g(x)) y2|{, T@$P./$% u(x)f(x) + v(x)g(x) = (f(x), g(x))  u(x)  v(x), N deg u(x) < deg µ g(x) (f(x), g(x))¶ , deg v(x) < deg µ f(x) (f(x), g(x))¶ . '(: DqI~ s(x), t(x) ∈ K[x] N s(x)f(x) + t(x)g(x) = (f(x), g(x)).  s(x) f(x) (f(x), g(x)) + t(x) g(x) (f(x), g(x)) = 1. (*) S s(x) = g(x) (f(x), g(x)) q(x) + u(x), -* u(x) = 0 k deg u(x) < deg g(x) (f(x), g(x)) . G v(x) = f(x) (f(x), g(x)) q(x) + t(x),  (*) u, u(x) f(x) (f(x), g(x)) + v(x) g(x) (f(x), g(x)) = 1. (**) 0 , f(x) (f(x), g(x))  g(x) (f(x), g(x)) y2|{,  u(x), v(x) 2`!{I~. ! deg u(x) < deg g(x) (f(x), g(x)) .  (**) u deg µ u(x) f(x) (f(x), g(x)) ¶ = deg µ v(x) g(x) (f(x), g(x)) ¶ , IJ deg v(x) < deg f(x) (f(x), g(x)) . % & 10–4 1. (f(x), m(x)) = 1, ^_: 4LI~ g(x), 2DqI~ h(x), N h(x)f(x) ≡ g(x) (mod m(x)). '(: 0 , Dq u(x), v(x) ∈ K[x], N u(x)f(x) + v(x)m(x) = 1.  g(x)u(x)f(x) + g(x)v(x)m(x) = g(x). ! g(x)u(x)f(x) ≡ g(x) (mod m(x)). · 6 ·

h(a)f(a)=g(r)(mod m(r)) 2.设m1(x),……,m2(x)为一组两两互素的多项式,证明:对任何的多项式f1(x),……,f3(x),都存在 多项式F(x),使 F(x)≡f(x)( mod mi(x),i=1,……,s 证明令M()=m(mn()…mn(x),B()=m1(x,则(B(),m(x)=1,m()|B() i≠j.存在h(x)使(习题1) hi()ri(a)=fi(e)(mod mi(a)) 令 F(a)=∑h(x)R(x) F(a)=>hi(r)R;(r)(mod mk(a)) ≡k(x)( mod mk(x) 3.设m(x)为复系数多项式,且m(0)≠0.证明:存在复系数多项式f(x),使 f2(x)≡x(modm(x) 证明:(a)首先证明对任意的a≠0,同余式 f2(x)≡x(mod(x-a)m) 有解设√a是a的任意一个平方根,则 (r-a)m=((va-va(va+vam=(vr-vam(v+va) =((x)V-g(x)(h(x)√G+g(x)=h2(x)x-9(x) 于是 92(a)=h(r)r(mod (a-a)m) 而h(a)√a+9(a)=(va+√am≠0,而h(a)√a-g(a)=(a-√am=0.因此g(a)h(a)≠0,从而 (h(x),(x-a厘m)=1,存在h1(x)∈K[x使h1(x)h(x)≡1(mod(x-a)m).于是 (hi(r)g(a -=r (mod (r-a)m) 取∫(x)=h1(x)9(x),则有 ≠a;对i≠j.则(x-a1) 两两互素.由(a),存在f(x)∈K],使 f2(x)≡x(mod(x-a)m) 由习题2,存在f(x)使 f(a)=fi(r)(mod (a-ai"o) 于是 f2(x)≡x(mod(x-a1)

S h(x) = g(x)u(x),  h(x)f(x) ≡ g(x) (mod m(x)). ∗2. m1(x), · · · , ms(x) .f]@@
I~, ^_: 4LI~ f1(x), · · · , fs(x), 2Dq I~ F(x), N F(x) ≡ fi(x) (mod mi(x)), i = 1, · · · , s. '(: S M(x) = m1(x)m2(x)· · · ms(x), Ri(x) = M(x) mi(x) .  (Ri(x), mi(x)) = 1, mj (x) | Ri(x), i 6= j. Dq hi(x) N (NO 1) hi(x)Ri(x) ≡ fi(x) (mod mi(x)) S F(x) = Xs i=1 hi(x)Ri(x),  F(x) ≡ Xs i=1 hi(x)Ri(x) (mod mk(x)) ≡ hk(x)Rk(x) (mod mk(x)) ≡ fk(x) (mod mk(x)). ∗3. m(x) .syI~, F m(0) 6= 0. ^_: DqsyI~ f(x), N f 2 (x) ≡ x (mod m(x)). '(: (a) !{^_45 a 6= 0, am f 2 (x) ≡ x (mod (x − a) m) $. √ a ! a 45fv/>x,  (x − a) m = ((√ x − √ a)(√ x + √ a))m = (√ x − √ a) m( √ x + √ a) m = (h(x) √ x − g(x))(h(x) √ x + g(x)) = h 2 (x)x − g 2 (x). ! g 2 (x) ≡ h 2 (x)x (mod (x − a) m) J h(a) √ a + g(a) = (√ a + √ a) m 6= 0, J h(a) √ a − g(a) = (√ a − √ a) m = 0, l g(a)h(a) 6= 0, IJ (h(x),(x − a) m) = 1, Dq h1(x) ∈ K[x] N h1(x)h(x) ≡ 1 (mod (x − a) m). ! (h1(x)g(x))2 ≡ x (mod (x − a) m) Q f(x) = h1(x)g(x), $ f 2 (x) ≡ x (mod (x − a) m). (b) m(x) = (x − a1) m1 (x − a2) m2 · · ·(x − as) ms , ai 6= aj  i 6= j.  (x − a1) m1 , · · · ,(x − as) ms @@
.  (a), Dq fi(x) ∈ K[x], N f 2 i (x) ≡ x (mod (x − ai) mi ). NO 2, Dq f(x) N f(x) ≡ fi(x) (mod (x − ai) mi ) ! f 2 (x) ≡ x (mod (x − ai) mi ) · 7 ·

由(x-a1)m1,…,(x-a,)m两两互素可得 f(x)≡x( 习题10-5 1.证明:g"(x)|f"(x)→g(x)|f(x) 证明设 f(x)=ap(x)2(x)…p(x) g(x)=bn(x)2(x)…p(x) 其中a,b∈K,p1(x),……,P(x)是两两互素的不可约多项式,且l1,k≥0,i=1,……,S.则 g(x)|f(x)←→k1≤l, 2.设f(x),9(x)∈K[x],且有分解式 ∫(x)=ap1(x)n2(x)…p(x),r;≥0,i=1,…,s g(x)=bn(x)n2(x)…z(x),t1≥0.i=1,…,s, 其中p1(x),……,Ps(x)是不同的首一不可约多项式证明: f(x),g(x)=mx(m)(x)2)(x) 明:令 (ri, ti), 则因r;≤mx,t≤mz,因此 f(x)|m(x),g(x)|m(x)=[f(x),g(x)]|m(x) 设s(x)∈K[]是f(x),9(x)的公倍式,则有 s(x)=n(x)2(x)…z(x)h(x),l1≤r,l≤t,(h(x),p(x)=1,i=1,…,s 于是 l;≥max(r;,t),i=1,…,s,=m(x)|s(x) 因此 f(x),g(x)=p11(x)p2(x)…p(x) 3.设f(x),9(x)∈K[]都是首一多项式,证明 If(a),g(a) f(ar)g(a) f(a), g(a)) f(x)=n1(x)p2(x)…p。(x),ra≥0, g(x)=n1(x)n2(x)…p(x),t≥0,i=1,…,s, 其中p1(x),…,P(x)是不同的首一不可约多项式令 mi= max(ri, ti), li= min(ri, ti), i=1

 (x − a1) m1 , · · · ,(x − as) ms @@
@O f 2 (x) ≡ x (mod m(x)). % & 10–5 1. ^_: g m(x) | f m(x) ⇐⇒ g(x) | f(x). '(: f(x) = ap l1 1 (x)p l2 2 (x)· · · p ls s (x), g(x) = bpk1 1 (x)p k2 2 (x)· · · p ks s (x), -* a, b ∈ K, p1(x), · · · , ps(x) !@@
`@I~, F li , ki > 0, i = 1, · · · , s.  g(x) | f(x) ⇐⇒ ki 6 li , i = 1, · · · , s ⇐⇒ mki 6 mli , i = 1, · · · , s ⇐⇒ g m(x) | f m(x). 2. f(x), g(x) ∈ K[x], F$ f(x) = ap r1 1 (x)p r2 2 (x)· · · p rs s (x), ri > 0, i = 1, · · · , s; g(x) = bpt1 1 (x)p t2 2 (x)· · · p ts s (x), ti > 0, i = 1, · · · , s, -* p1(x), · · · , ps(x) !`a!f`@I~. ^_: [f(x), g(x)] = p max(r1,t1) 1 (x)p max(r2,t2) 2 (x)· · · p max(rs,ts) s (x). '(: S mi = max(ri , ti), i = 1, · · · , s. m(x) = p m1 1 (x)p m2 2 (x)· · · p ms s (x),  ri 6 mi , ti 6 mi , l f(x) | m(x), g(x) | m(x) =⇒ [f(x), g(x)] | m(x). s(x) ∈ K[x] ! f(x), g(x) , $ s(x) = p l1 1 (x)p l2 2 (x)· · · p ls s (x)h(x), li 6 ri , li 6 ti , (h(x), pi(x)) = 1, i = 1, · · · , s. ! li > max(ri , ti), i = 1, · · · , s, =⇒ m(x) | s(x). l [f(x), g(x)] = p m1 1 (x)p m2 2 (x)· · · p ms s (x). 3. f(x), g(x) ∈ K[x] 2!!fI~, ^_: [f(x), g(x)] = f(x)g(x) (f(x), g(x)). '(: f(x) = p r1 1 (x)p r2 2 (x)· · · p rs s (x), ri > 0, i = 1, · · · , s; g(x) = p t1 1 (x)p t2 2 (x)· · · p ts s (x), ti > 0, i = 1, · · · , s, -* p1(x), · · · , ps(x) !`a!f`@I~. S mi = max(ri , ti), li = min(ri , ti), i = 1, · · · , s. · 8 ·

f(x)g(x)=p1+t1(x)p2+2(x)…p+(x) (f(x,g(x)=n(x)2(x)…p(x) 由于r2+t1-l=m,i=1,……,s.因此 (.92)=n"()p2(n)…p()=U(x),g( 4.求下列多项式的最小公倍式 (1)f(x)=x4-4x3+1,g(x)=x3-3x2+1 (2)f(x)=x4-x-1+i.g(x)=x2+1 解:(1)由于(f(x),g(x)=1,[f(x),9(x)]=f(x)9(x)=x7-7x6+12x5+x4-3x3-3x2+1 (2)由于(f(x),9(x)=x-i,f(x),9(x)]=f(x)(x+i)=x5+i4-x2-x-(1+i) 5.设p(x)是次数大于零的多项式证明:如果对于任何多项式f(x),9(x),由p(x)|f(x)9(x)可以 推出p(x)|f(x)或者p(x)|g(x),则p(x)是不可约多项式 证明:若p(x)可约,则存在次数小于p(x)的非常数多项式f(x),9(x)使p(x)=f(x)9(x).从而 p(x)lf(x)g(x).但因 deg f(a)< degp(a), deg g(a)< deg p(a), p(x)f(x),p(x)g(x),与假设矛盾,因此p(x)不可约 6.证明:次数大于0的首一多项式f(x)是某一不可约多项式的方幂的充分必要条件是,对任意的 多项式g(x)必有(f(x),9(x)=1,或者对某一正整数m,f(x)g"(x) 证明:(→)设f(x)=p(x),其中p(x)不可约,则若9(x)∈K[al满足p(x)|g(x),有 如p(x)g(x),则(p(x),g(x)=1,从而(pm(x),g(x)=1,即(f(x),9(x)=1 (←)设p(x)是f(x)的一个首一不可约因子,则(p(x),f(x))=p(x),从而存在某个正整数m,使 ∫(x)|p"(x),这说明p(x)是f(x)的唯一不可约因子.所以f(x)=cp(x).又因f(x),p(x)的首项系数都 是1,故c=1.从而f(x)=p(x) 7.证明:次数大于0的首一多项式f(x)是某一不可约多项式的方幂的充分必要条件是,对任意的 多项式9(x),h(x),由f(x)|g(x)h(x)可以推出f(x)|g(x),或者对某一正整数m,f(x)|hm(x) 证明:(→)设f(x)=p(x),其中p(x)是首一不可约多项式,则由f(x)|g(x)h(x),可得p(x) g(x)h(x),从而p(x)|g(x)或p(x)|h(x).于是f(x)=p(x)|gm(x)或∫(x)=p(x)|h(x) ()设p(x)是f(x)的一个首一不可约因子,则f(x)=p(x)f1(x).从而f(x)|p(x)f1(x).而 ∫(x)十f1(x),从而存在某个正整数m,使f(x)|pm(x),这说明p(x)是∫(x)的唯一不可约因子.所以 f(x)=cp(x).又因f(x),p(x)的首项系数都是1,故c=1.从而f(x)=p(x) 习题106 1.判别下列有理系数多项式有无重因式,若有,则求出重因式 (1)f(x)=x5-10x3-20x2-15x-4 (2)f(x)=x4-4x3+16x-16; (3)f(x)=x5-6r4+16x3-24x2+20x-8 1)f(x)=x°-15x4+8x3+51x2-72x+2

f(x)g(x) = p r1+t1 1 (x)p r2+t2 2 (x)· · · p rs+ts s (x), (f(x), g(x)) = p l1 1 (x)p l2 2 (x)· · · p ls s (x),  ri + ti − li = mi , i = 1, · · · , s. l f(x)g(x) (f(x), g(x)) = p m1 1 (x)p m2 2 (x)· · · p ms s (x) = [f(x), g(x)]. 4. L,I~^": (1) f(x) = x 4 − 4x 3 + 1, g(x) = x 3 − 3x 2 + 1; (2) f(x) = x 4 − x − 1 + i. g(x) = x 2 + 1. ): (1)  (f(x), g(x)) = 1, [f(x), g(x)] = f(x)g(x) = x 7 − 7x 6 + 12x 5 + x 4 − 3x 3 − 3x 2 + 1. (2)  (f(x), g(x)) = x − i, [f(x), g(x)] = f(x)(x + i) = x 5 + ix 4 − x 2 − x − (1 + i). 5. p(x) !y|{I~. ^_: 454LI~ f(x), g(x),  p(x) | f(x)g(x) @ 1 p(x) | f(x) k2 p(x) | g(x),  p(x) !`@I~. '(:  p(x) @, Dqy" p(x) z`yI~ f(x), g(x) N p(x) = f(x)g(x). IJ p(x) | f(x)g(x). A deg f(x) 'CDEFG!, 45 I~ g(x) D$ (f(x), g(x)) = 1, k23fW y m, f(x) | g m(x). '(: (⇒) f(x) = p m(x), -* p(x) `@,  g(x) ∈ K[x]  p(x) | g(x), $ f(x) = p m(x) | g m(x). 4 p(x) - g(x),  (p(x), g(x)) = 1, IJ (p m(x), g(x)) = 1, K (f(x), g(x)) = 1. (⇐) p(x) ! f(x) fv!f`@ ,  (p(x), f(x)) = p(x), IJDq3vW y m, N f(x) | p m(x), k_ p(x) ! f(x) 4f`@ .  f(x) = cpr (x). C f(x), p(x) !~sy2 ! 1, 3 c = 1. IJ f(x) = p r (x). ∗7. ^_: y| 0 !fI~ f(x) !3f`@I~>'CDEFG!, 45 I~ g(x), h(x),  f(x) | g(x)h(x) @1 f(x) | g(x), k23fW y m, f(x) | h m(x). '(: (⇒) f(x) = p m(x), -* p(x) !!f`@I~,  f(x) | g(x)h(x), @O p(x) | g(x)h(x), IJ p(x) | g(x) k p(x) | h(x). ! f(x) = p m(x) | g m(x) k f(x) = p m(x) | h m(x). (⇐) p(x) ! f(x) fv!f`@ ,  f(x) = p(x)f1(x). IJ f(x) | p(x)f1(x). J f(x) - f1(x), IJDq3vW y m, N f(x) | p m(x), k_ p(x) ! f(x) 4f`@ .  f(x) = cpr (x). C f(x), p(x) !~sy2! 1, 3 c = 1. IJ f(x) = p r (x). % & 10–6 1. },$￾syI~$ w, $, Lw: (1) f(x) = x 5 − 10x 3 − 20x 2 − 15x − 4; (2) f(x) = x 4 − 4x 3 + 16x − 16; (3) f(x) = x 5 − 6x 4 + 16x 3 − 24x 2 + 20x − 8; (4) f(x) = x 6 − 15x 4 + 8x 3 + 51x 2 − 72x + 27. · 9 ·

解:(1)x+1,4重 (2)x-2,3重 (3)x2-2x+2,2重 (4)x+3,2重,x-1,3重 2.a,b应满足什么条件,下列多项式有重因式? 1)f(x)=x3 (2)f(x)=x4+4ax+b 解:(1)当a=b=0有3重因式x,当4a3=-b2且a≠0,有2重因式2ax+b (2)当a=b=0有4重因式x,当27a4=b3且a≠0,有2重因式3ax+b 3.设p(x)首f(x)的k重因式,能否说p(x)首f(x)的k+1重因式,为什么? 解:不能.因为又可能f(x)任一重因式都不首f(x)的因式例如f(x)=x4-1,f(x)=4x3 4.证明:如果(f(x),f"(x))=1,那么,f(x)的重因式都首f(x)的二重因式 证明:由于(f(x),f"(x)=1,f(x)的任一因式都不首f"(x)的因式设p(x)首f(x)的重因式,则 p(x)|f(x),于首p(x)十f"(x),说明p(x)首∫(x)的单因式故p(x)首f(x)的二重因式 5.证明:K中不可约多项式p(x)首f(x)∈K的k(k≥1)重因式的充分必要条件首p(x)首 f(x),f"(x),…,f(k-1)(x)的因式但不首f(k)(x)的因式 证明:(→)对k用归纳法.当k=1时结论显然成立.现设结论对k-1成立设p(x)首f(x)的 重因式则f(x)=p(x)9(x),其中(p(x),g(x)=1.则 f(x)=kp-1(x)9(x)+p(x)g(x)=p-1(x)(kg(x)+p(x)g(x) 由(p(x),g(x)=1可得(p(x),kg(x)+p(x)g(x)=1,因此p(x)首f(x)的k-1重因式根据归纳假设 p(x)首f(x),…,fk-1)(x)的因式,但不首f()(x)的因式而p(x)首f(x)的因式首已知的 (=)如p(x)首f(x),f(x),…,f(k-1)(x)的因式但不首f()(x)的因式,则p(x)首f(-1)(x)的 重因式进而,p(x)首fk-2)(x)的二重因式依次类推可知p(x)首f(x)的k重因式 6.试求多项式x19+1除以(x-1)2所得余式 解:设x19+1=(x-1)2q(x)+ax+b,则两边求导后得 1999192(x-1)q(x)+(x-1)2q(x) 以x=1代入上两式,得 a=1999,b=-1997 故所求余式为1999x-199 习题107 1.求下列多项式的公共根: (1)f(x)=x4+2x2+9,g(x)=x4-4x3+4x2-9 (2)f(x)=x3+2x2+2x+1,g(x)=x4+x3+2x2+x+1 解:(1)1+√2,1-√2 2.如果(x-1)2|Ax4+Bx2+1,求A,B 解:A=1,B 3.已知x4-3x3+6x2+ax+b能被x2-1整除,求a,b

): (1) x + 1, 4 w. (2) x − 2, 3 w. (3) x 2 − 2x + 2, 2 w. (4) x + 3, 2 w, x − 1, 3 w. 2. a, b 5"#FG, ,I~$w? (1) f(x) = x 3 + 3ax + b; (2) f(x) = x 4 + 4ax + b. ): (1) P a = b = 0 $ 3 w x, P 4a 3 = −b 2 F a 6= 0, $ 2 w 2ax + b. (2) P a = b = 0 $ 4 w x, P 27a 4 = b 3 F a 6= 0, $ 2 w 3ax + b. 3. p(x) ! f 0 (x)  k w, U p(x) ! f(x)  k + 1 w, ."#? ): `U. .C@U f 0 (x) 4fw2`! f(x) . ?4 f(x) = x 4 − 1, f 0 (x) = 4x 3 . 4. ^_: 45 (f 0 (x), f00(x)) = 1, [#, f(x) w2! f(x) w. '(:  (f 0 (x), f00(x)) = 1, f 0 (x) 4f2`! f 00(x) . p(x) ! f(x) w,  p(x) | f 0 (x), ! p(x) - f 00(x), _ p(x) ! f 0 (x) , 3 p(x) ! f(x) w. 5. ^_: K[x] *`@I~ p(x) ! f(x) ∈ K[x]  k (k > 1) wCDEFG! p(x) ! f(x), f0 (x), · · · , f(k−1)(x) , A`! f (k) (x) . '(: (⇒)  k X67q. P k = 1 S898:;. 9 89 k − 1 :;. p(x) ! f(x)  k w,  f(x) = p k (x)g(x), -* (p(x), g(x)) = 1.  f 0 (x) = kpk−1 (x)g(x) + p k (x)g 0 (x) = p k−1 (x)(kg(x) + p(x)g 0 (x)).  (p(x), g(x)) = 1 @O (p(x), kg(x) + p(x)g 0 (x)) = 1, l p(x) ! f 0 (x)  k − 1 w. x-670 , p(x) ! f 0 (x), · · · , f(k−1)(x) , A`! f (k) (x) . J p(x) ! f(x) !tu. (⇐) 4 p(x) ! f(x), f0 (x), · · · , f(k−1)(x) , A`! f (k) (x) ,  p(x) ! f (k−1)(x) f w, :J, p(x) ! f (k−2)(x) w, ;<1, @u p(x) ! f(x)  k w. 6. <LI~ x 1999 + 1 " (x − 1)2 Om. ): x 1999 + 1 = (x − 1)2 q(x) + ax + b, @RL=O 1999x 1998 = 2(x − 1)q(x) + (x − 1)2 q(x) + a.  x = 1 b @, O a = 1999, b = −1997. 3Lm. 1999x − 1997. % & 10–7 1. L,I~x: (1) f(x) = x 4 + 2x 2 + 9, g(x) = x 4 − 4x 3 + 4x 2 − 9; (2) f(x) = x 3 + 2x 2 + 2x + 1, g(x) = x 4 + x 3 + 2x 2 + x + 1. ): (1) 1 + √ 2i, 1 − √ 2i. (2) −1 + √ 3i 2 , −1 − √ 3i 2 . 2. 45 (x − 1)2 | Ax4 + Bx2 + 1, L A, B. ): A = 1, B = −2. 3. tu x 4 − 3x 3 + 6x 2 + ax + b U) x 2 − 1 ", L a, b. · 10 ·