2010年寒假數(shù)學(xué)集訓(xùn)六導(dǎo)學(xué)jx6-1多項(xiàng)式的運(yùn)算與基本定理

多項(xiàng)式的運(yùn)算與基本定問(wèn)題引

1問(wèn)題

AB

x(x1)(x

x x1解析

AB

x(x1)(xx(x1)(x

x x 1A(x1)(x2)Bx(x2)Cx(x x0，得12A,A2 x x

B1得12CC12 2

1 x(x1)(x

x

x問(wèn)題 設(shè)a、b、c是三個(gè)不同的實(shí)數(shù)，P(x)是實(shí)系數(shù)多項(xiàng)式，已知：①P(x)除以x所得余式為aP(xxb所得余式為bP(xxc所得余式為求多項(xiàng)式P(x)除以(xa)(xb)(xc)所得余式解法一

P(xxa除，余數(shù)為aP(a)aP(b)bP(c)多項(xiàng)式T(xP(xxT(aP(aa0T(b0，T(c0T含因式(xa)(xb)(xcT(x)P(xxP(x除以(xa)(xb)(xc的余x。解法二P(x)xa)(xb)(xc)g(xr(x，其中r(x)0或degr(x)r(aar(bbr(cc，由多項(xiàng)式恒等定理有r(x)x基本知一、形如f(x)axn xn1 ax

0x的一元n nf(x的次數(shù)，記為degf(xai(0inx的iaxna 當(dāng)degf(x0且a00f(x稱為零次多項(xiàng)式。數(shù)0當(dāng)系數(shù)an, 同一數(shù)系內(nèi)的若干多項(xiàng)式的和、差、積還是這一數(shù)系內(nèi)的多項(xiàng)式，且deg(f(x)max|degf(xdegg(x|;deg(f(xg(x))degf(xdegg(x);其中f(x)g(x)均是非零多項(xiàng)式。若多項(xiàng)式f(x)不能表示成兩個(gè)次數(shù)比f(wàn)(x)f(x二、若f(x)g(x)g(x)q(xr(xf(x)g(xq(xr(x)。其中degr(xdegg(x)r(x)0。q(xr(xf(xg(x)r(x)0g(xf(xg(xf(xg(x|fg(x|f(xf(x|g(x)存在非零常數(shù)Cf(x)Cg(x

g(x)|f(x)

h(x)|g(x)

h(x)|f（3）g(x)|f(xg(x)|h(x)g(x)|g(x)[u(x)f(xv(x)h(x)]，其中u(x)定理 若有n1個(gè)不同的x值使n次多項(xiàng)式f(x)與g(x)的值相同，則f(x)g(x)2定理 多項(xiàng)式f(x)除以xa所得的余數(shù)等于ff(xxa)g(xf3定理 多項(xiàng)式f(x)有一個(gè)因式xa的充要條件是f(a)f(xf(xxaf(xa、b為不同整數(shù)，則(ab|f(af5

f(xpxqp0)

f p(xa1)(xa2) (xan)(a0a1)(a0a2) (a0an)(xa1)(xa2) (xan)(a0a1)(a0a2) (a0an)(xa0)(xa2) (xan) (a1anf(x)

f(a0

f (xa0)(x(xa0)(xa1) (xan1)(ana0)(ana1) (anan1)說(shuō)明：Lagrange插值告訴我們，只要已知n次多項(xiàng)式在n1個(gè)點(diǎn)處的值，就能直接確定這個(gè)n次多項(xiàng)式妙題巧a0 解析設(shè)f(x)axna0 g(x)bxn h(x)f(x)g(x) xnm xmn1 如果h(x)不是本原的，也就是說(shuō)h(x)的系數(shù)dnm,dnm1, ,do有一異于1的公因子，那么就有一個(gè)素?cái)?shù)能整除h(x)的每一個(gè)系數(shù)。f(xpf(x的每一個(gè)系數(shù)，令aip除的系數(shù)，即p|a0,p| ,p|ai1,p同樣地，

pp|bop| p|bi1p|bi我們來(lái)看h(x)的系數(shù)dijdijaibjai1bj1ai2bj2p|dij,p整除上式右端除以aibjp|aibj明了h(x)例 （1）證明：多項(xiàng)式f(x)為偶函數(shù)的充要條件是f(x)沒(méi)有奇次項(xiàng)（2）證明：(1xx2x3 x99x100)(1xx2x3 合并同類項(xiàng)后，無(wú)x的奇次項(xiàng)。解析（1）設(shè)f(x)aax1ax2 axn為偶函數(shù)，則f(x)f(x)aax1a ax3 1)naxn f(xaax2ax4 f(x （2）令g(x)(1xx2x3 x99x100)(1xx2x3 g(x)(1xx2x3 x99x100)(1xx2x3 x99x100)g(x)g(x為偶函數(shù)。g(x（1） (((x2)22)2 (((x2)22)23pn(x

，其中nN*，試求P(xx2項(xiàng)nn解析

pn(x)(

n2)2,P(xx2xna,b,c

(x)(p(x)2)2

ax2bx

(c2)2 bn12bn(cn b22a

又a11b14c1c 4b

42n

1(42n14n1)例 已知f(x)是x的n次多項(xiàng)式(n0)，且對(duì)任意的實(shí)數(shù)x滿足8f(x3)x6f(2x)2f(x2)12

f(x解析

f(x的最高次項(xiàng)為axn(a08f(x3x6f(2x2f(x2 為8ax3n2naxn6 因?yàn)?n3n3nnnn8an2

所以n f(xax3ax2a 0x0，代入8f(x3x6f(2x2f(x2120，得a0 f(xax3ax2ax 4ax82ax7(8a22a)x62ax48ax32ax2 4a20,2a10,8a222a3a31,a20,a10,a0f(xx35f(xx4x3x2x1f(x5f(x解 x51(x1)(x4x3x2xf(x)|(x5f(x5x51yfyy1f(1fyqyy15f(x5)q(x5)(x1)f(x)f(x5f(x點(diǎn)評(píng)：當(dāng)n是奇數(shù)時(shí)，xn1(x1)(xn1xn2 x1)xn1x1)(xn1xn2 x1；當(dāng)nxxn1(x1)(xn1xx

xn1(x1)(xn1xn2

但xn1xx1x例 求證：f(x)x9999x8888 x11111能被g(x)x9x8 x1整解析f(xg(x)x11111)(x9x11111)(x9x8 x(x9999x9)(x8888x8) (x1111x9[(x10)9991]x8[(x10)8881] x101x1)g(xf(xg(xg(x)點(diǎn)評(píng):g(x|f(xg(xg(x|f(x來(lái)轉(zhuǎn)化。并注意到(x10nxx101x101x9x例 已知：f(x) ))x，P(x)x，求證：P(x)x|f解析：顯然xy|(P(x)P(y))，則 f(xPn(x[P(P(x))P(x)][P(x)P(xx|P(xxP(xx|P(P(xP(xP(P(xP(x)P2xP(xxy|P(xPy從而P(x)x整除P(x)x，P2(x)P(x)，P3(x)P2(x), 所以P(x)x|f(x)點(diǎn)評(píng)：從簡(jiǎn)單情形入手，當(dāng)n2f(xP(P(xx把f

分成能被P(x

[P(P(xP(x)]、P(x

，只需要P(xx|P(P(xP(xxy|P(xPy例 已知f(x)ax3bx2cxd滿1f(12,1f(132 f(4)解析：利用Lagrange插值，f(4)

(41)(4(11)(1(41)(41)(43)f(2)(4(21)(21)(2 1f(1)5f(1)5f(2)15f 2 154 3即4

f(4)點(diǎn)評(píng)：利用Lagrange插值 可估計(jì)多項(xiàng)式f(x)在xx0處的取值情況。本題f(4)范圍例 設(shè)P(x)為2n次多項(xiàng)式，滿足P(0P(2 P(2n0P(1P(3 P(2n12P(2n1n,解 令f(x)P(x)1，于是f(k)(1)k1,k,(xx1)(xx2 (xx(xx1)(xx2 (xx2n(x0x1)(x0x1 (x0x1((xx0)(xx1 (xxk1)(xxk1 (xx2n(xkx0)(xkx1 (xkxk1)(xkxk1(xkx2nf(k) (xx(xx0)(xx1 (xx2n1(x2nx0)(x2nx1(x2nx2n1,xkk(k,2n(2n1) 2n(2n1) f(2n1)(2n1)(2n1)2nk(k1)1(1)(2)(2nk2)(2nk) 1(k(2n1)2(2n1)2n 22n(2n1) 1(2n)!(1)(1)k1(2n1)2n (2nk2)(2nk)! (2n(2n(2n1)2n(2nk2)k(2n

]

C2n1) [22n11]1P(2n130f(2n1)P(2n1)1311解得nP(xP(0P(2P(40P(1P(32P(x)

x(x2)(x3)(x

x(x1)(x2)(x 2x416x340x232 點(diǎn)評(píng)：直線運(yùn)用Lagrange插值求P(x)，雖有P(0)P(2) P(2n)0,，使插值中項(xiàng)數(shù)減少，但運(yùn)算量并未從根本上降低，重新構(gòu)造多項(xiàng)式f(x)P(x)1后，有f(k1)k1(k0,1 模擬試題6-f(xax3cx滿足1

f(2)4，則f(3)的取值范圍 16f(3)

7

f(3)

C16

f(3)20

23f(3) 設(shè)(1xx2)naaxax2 ax2n為x的一個(gè)恒等式 記a2a4 a2n，則 2n

3nC2

3nD 設(shè)abx2x1ax3bx21的一個(gè)因式，則a2bx1002x551x21f(xcosxsin)ncosnxsinnx21f(xg(xh(xf2x)xg2xxh2f(x)g(x)h(x)x43x22x3已知aZx633x20x2xa整除，求aP(xx4ax3bx2cxda,bcdP(1668P(2)1336P(320041[P(11P(7)]8設(shè)P(x)是一個(gè)n次多項(xiàng)式(n0)，且對(duì)k0,1,2, ,n。有P(k)P(n

kk已知數(shù)列a0,a1,a2, ,滿足a0a1且ai1ai12ai(i1,2,3, 然數(shù)n，P(x)aC0(1x)naC1x(1x)n1aC2x2(1x)n2 aCn1xn1(1x)aCnxn0 1 2

nf(x2f2xfP為ABCn

3 3 [

n

nn，求證Ck2k

2]Cn

，其中C0 ]表

k

完全找到整系數(shù)不可約多項(xiàng)式的一種新大家知道，有限域上的不可約多項(xiàng)式的研究一直受到數(shù)學(xué)界、編碼與領(lǐng)域的普遍關(guān)即使判定一個(gè)給定的多項(xiàng)式不可約，也是比較而繁雜的問(wèn)題。反過(guò)來(lái)，找到滿足要求的定理1（廣義算術(shù)基本定理）r,在不記因素順序的條件下，可以唯一地表示為，其中ai(i=1,2,?,n)為整數(shù),2,3,5,?,pn為前n+1n證明：（1）當(dāng)r=1時(shí)，當(dāng)且僅當(dāng)ai=0(i=0,1,2n)時(shí)，才有1=2a03a15a2nnr=2a03a15a2pan，其中ai(i=0,1,2n)n 滿足r=a b從而 ,|a

?

(????0,????1,?,??????<注意到 故一切pil(l=0,1,2,?,n)皆為互不相同的素?cái)?shù)。經(jīng)過(guò)對(duì)pil(l=0,1,2,?,n)位置調(diào)整，得到n+n表示整系數(shù)多項(xiàng)式a0+a1x+a2x2+?+anxn(2)，我們稱這種對(duì)應(yīng)關(guān)系為式數(shù)對(duì)應(yīng)法則。{2}?22={x2}?{x3}?{3}?23={2x}?32={1+x2}?2151={x4}?{x5}?令1的元素唯一地對(duì)應(yīng)Q+中的元素，反之，Q+中的元素可以唯一地表示1的元（2）nn證明：（1）對(duì)任一個(gè)整系數(shù)多項(xiàng)式a0+a1x+a2x2+