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
% &'( ) % & 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, J y. 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 ·