最高公因式与最低公倍式.doc

肆芃薆袀腿膆蒂衿袈莂莈袈羀膅蚆袇肃莀薂袆膅膃蒈羅袅莈莄羄羇膁蚃羄腿莇虿羃节艿薅羂羁蒅蒁薈肄芈莇薈膆蒃蚆薇袆芆薂蚆羈蒁蒇蚅肀芄莃蚄芃肇螂蚃羂莃蚈蚂肅膅薄蚂膇莁蒀蚁袆膄莆蚀罿荿蚅蝿肁膂薁螈膃莇蒇螇袃膀蒃螆肅蒆荿螆膈艿蚇螅袇蒄薃螄羀芇葿螃肂蒂莅袂膄芅蚄袁袄肈薀袀肆芃薆袀腿膆蒂衿袈莂莈袈羀膅蚆袇肃莀薂袆膅膃蒈羅袅莈莄羄羇膁蚃羄腿莇虿羃节艿薅羂羁蒅蒁薈肄芈莇薈膆蒃蚆薇袆芆薂蚆羈蒁蒇蚅肀芄莃蚄芃肇螂蚃羂莃蚈蚂肅膅薄蚂膇莁蒀蚁袆膄莆蚀罿荿蚅蝿肁膂薁螈膃莇蒇螇袃膀蒃螆肅蒆荿螆膈艿蚇螅袇蒄薃螄羀芇葿螃肂蒂莅袂膄芅蚄袁袄肈薀袀肆芃薆袀腿膆蒂衿袈莂莈袈羀膅蚆袇肃莀薂袆膅膃蒈羅袅莈莄羄羇膁蚃羄腿莇虿羃节艿薅羂羁蒅蒁薈肄芈莇薈膆蒃蚆薇袆芆薂蚆羈蒁蒇蚅肀芄莃蚄芃肇螂蚃羂莃蚈蚂肅膅薄蚂膇莁蒀蚁袆膄莆蚀罿荿蚅蝿肁膂薁螈膃莇蒇螇袃膀蒃螆肅蒆荿螆膈艿蚇螅袇蒄薃螄羀芇葿螃肂蒂莅袂膄芅蚄袁袄肈薀袀肆芃薆袀腿膆蒂衿袈莂莈袈羀膅蚆袇肃莀薂袆膅膃蒈羅袅莈莄羄羇膁蚃羄腿莇虿 最高公因式與最低公倍式主題1：因式與倍式1.因式、倍式：設、為二多項式，且，若存在使，則稱為之因式，為之倍式，以。2.公因式：若多項式同時是與之因式，則稱為與之公因式。(兩多項式的公因式有無窮多個。)3.互質：若多項式與除常數多項式外，沒有其他的公因式，則稱它們互質。4.最高公因式（H.C.F）：設,為兩多項式，若是它們的公因式中次數最高的，則就稱為,之最高公因式。常以或（,）表示。(最高公因式不唯一，但只差常數倍)5.公倍式：設,都是非零多項式，若同時是, 之倍式，則稱為,之公倍式。6.最低公倍式（L.C.M）：,的公倍式中，次數最小的，稱為其最低公倍式，常以，表示或以表示。(最低公倍式不唯一，但只差常數倍)7.8.9.牛頓定理【一次因式檢驗法】：為整係數多項式；，若為之因式則且。重要範例1.若deg f (x) = 3，deg g (x) = 3，選出下列正確的敘述：(A) deg(f (x) + g (x) = 6(B)deg(f (x) + g (x) = 3(C) deg(f (x)g (x) = 6(D) deg(f (x)g (x) = 9(E)若f (x)與g (x)之最高公因式為一次式，則f (x)與g (x)之最低公倍式為五次式【註】deg f (x)表示多項式f (x)的次數。 【解答】(C)(E)【詳解】(A)(B)錯。deg(f (x) + g (x) 3(C)(D)deg (f (x)g(x) = degf (x) + deg g(x) = 3 + 3 = 6(C)對，(D)錯(E)對。由f (x)g (x) = k (f (x)，g (x)f (x)，g (x)deg(f (x)g (x) = deg(f (x)，g (x) + degf (x)，g (x)6 = 1 + degf (x)，g (x)故degf (x)，g (x) = 5隨堂練習.設deg f (x) = 5，deg g(x) = 3，則下列敘述何者正確？(A) deg (HCF) deg (LCM)(B) deg (HCF) + deg (LCM) = 8(C) deg (HCF) 最大是3(D) deg (LCM) 最小是5(E) deg (LCM) 最大是7。 【解答】(A)(B)(C)(D)【詳解】(A)HCF | LCMdeg (HCF) deg (LCM)(B)deg f (x) + deg g(x) = deg (HCF) + deg (LCM) = 8(C)(D) deg (HCF)最大 = deg f (x)，deg g(x)中之較小者deg (LCM)最小(E) deg (LCM)最大 = deg f (x) + deg g(x) = 82.若f (x) = 2x2 + ax + b，g(x) = 3x2 + cx + d的HCF為x + 1，LCM為6x3 + 17x2 + 14x + e，a，b，c，d，eR，則(A) a = 5(B) b = 3(C) c = 4(D) d = 1(E) e = 3。【解答】(A)(B)(C)(D)(E)【詳解】1 HCF | LCMx + 1 | 6x3 + 17x2 + 14x + ee = 32 LCM = 6x3 + 17x2 + 14x + 3 = (x + 1)(2x + 3) (3x + 1)3 f (x) = 2x2 + ax + b = (x + 1) (2x + 3)a = 5，b = 3又g(x) = 3x2 + cx + d = (x + 1)(3x + 1)c = 4，d = 13.a為整數，f (x) = x3 - x2 + 4x + a - 7，g(x) = 2x3 + 5x2 - 7x + 2a - 6，已知f (x)和g(x)的最高公因式為一次式，求a之值。 【解答】3【詳解】f (x) = x3 - x2 + 4x + a - 7，g(x) = 2x3 + 5x2 - 7x + 2a - 6設f (x)與g(x)的最高公因式為d(x)d(x) | g(x) - 2 f (x)d(x) | 7x2 -15x + 8d(x) | (7x - 8)(x - 1)d(x) = 7x - 8或x -1aZd(x) = x -1因d(x) | f (x)，d(x) | g(x)x -1 | f (x)，x -1 | g(x)f (1) = 0，g(1) = 01 -1 + 4 + a - 7 = 0a = 34.設a為整數，若f (x) = x3 + x2 - 4x + (a - 7)與g(x) = 2x3 - 7x2 + 7x + (2a - 8)的最高公因式為一次式，則a的值為。 【解答】3【詳解】已知f (x) = x3 + x2 - 4x + (a - 7)與g(x) = 2x3 - 7x2 + 7x + (2a - 8)之HCF為一次式，設為d(x)，則d(x) | 2f (x) - g(x)d(x) | 3 (3x2 - 5x - 2)d(x) | 3(x - 2)(3x + 1)d(x) = x - 2，或d(x) = 3x + 1(1)若d(x) = x - 2，則x - 2 | f (x)f (2) = 08 + 4 - 8 + (a - 7) = 0a = 3(2)若d(x) = 3x + 1，則3x + 1 | f (x)f () = 0非整數，不合，故所求a之值為35.設k為實數，f (x) = x3 - 2x2 - 5x + 6，g(x) = x3 + k2x2 + 2kx -16。(1) f (x)與g(x)有一次式的最高公因式時，k之值為何？(2) f (x)與g(x)有二次式的最高公因式時，k之值為何？【解答】(1) k = - 5或 - 2(2) k = 3【詳解】f (x) =，先將f (x)因式分解得f (x) = (x -1)(x + 2)(x - 3)x -1 | g(x)g(1) = 0k 2 + 2k -15 = 0k = 3或 - 5x + 2 | g(x)g( - 2) = 04k 2 - 4k - 24 = 0k = 3或 - 2x - 3 | g(x)g(3) = 09k 2 + 6k + 11 = 0k無實數解(1) f (x)與g(x)有一次式的HCFk = - 5時，HCF為x -1；k = - 2時，HCF為x + 2(2) k = 3時，HCF為二次式 (x - 1)(x + 2)隨堂練習.設f (x) = x3 + ax2 - 5x + (a - 7)與g(x) = x2 + ax - 2的HCF為一次式，則a = 。（a為整數） 【解答】1【詳解】設f (x)，g(x)的HCF為d(x)則d(x) | f (x) - xg(x)d(x) | - 3x + (a - 7)d(x) | - 3 (x -)d(x) | g(x)g() = 04a2 - 35a + 31 = 0(a -1)(4a - 31) = 0a = 1或a =，但a為整數a = 16.若x2 + px - 5與x3 + px - 5的最大公因式為一次式，求p =。 【解答】4【詳解】d (x) | (x3 + px - 5)且d (x) | (x2 + px - 5)，則d (x) | (x3 + px - 5) - (x2 + px - 5)即d (x) | x2 (x - 1)，d (x) = x，x2均不合，取d (x) = x -1(x - 1) | (x2 + px - 5)，即1 + p - 5 = 0，p = 4隨堂練習.若多項式f (x) = x2 + px + 6，g(x) = x3 + px + 6的LCM為四次式，則常數p之值為。【解答】- 7【詳解】1 f (x)，g(x)之HCF為一次式2 HCF | f (x)，HCF | g(x)HCF | g(x) - f (x) = x3 - x2 = x2 (x -1)3 但f (0) = 6 0HCF = x -14 由f (1) = 0p = - 77.已知二多項式x3 + ax2 + 2x + 4與2x3 + x2 + a2 x + 2有一個二次公因式，則a的值為。【解答】2【詳解】設f (x) = x3 + ax2 + 2x + 4，g(x) = 2x3 + x2 + a2x + 2已知f (x)與g(x)有一個二次公因式，設為d(x)，則d(x) | 2f (x) - g(x)d(x) | (2a -1) x2 + (4 - a2) x + 6又d(x) | f (x) - 2g(x)d(x) | - x 3x2 + (2 - a)x + 2(a2 -1)因x非f (x)與g(x)的因式，所以d(x) | 3x2 + (2 - a)x + 2(a2 -1)由，知若a = 2，則，a = 2為其解若a 2，則2a -1 = 6 + 3aa = - 7但 - 5 =a = - 7不合8.設f (x) = x3 + ax2 - 8x + 5，g (x) = 2x3 + bx2 - 7x - 5，若f (x)與g (x)的最高公因式為二次式，求(1)數對(a，b) =。(2)最高公因式為。 【解答】(1) (2，7)(2) x2 + 3x - 5【詳解】設d (x) = (f (x)，g (x)，且deg d (x) = 2則d (x) | 2f (x) - g (x)d (x) | (2a - b)x2 - 9x + 15d (x) | f (x) + g (x)d (x) | x 3x2 + (a + b)x - 15由d (x) = (2a - b)x2 - 9x + 15由d (x) = 3x2 + (a + b)x - 15（f (0) 0，g (0) 0）=得數對(a，b) = (2，7)，HCF = d (x) = x2 + 3x - 5隨堂練習.兩多項式f (x) = x3 + ax2 - 4x + 2與g(x) = x3 + bx2 - 2的最高公因式為二次式，則數對(a，b) = 。 【解答】(1，3)【詳解】設f (x)與g(x)的HCF為d(x)，則d(x) | f (x) - g(x)d(x) | (x3 + ax2 - 4x + 2) - (x3 + bx2 - 2)d(x) | (a - b) x2 - 4x + 4d(x) | f (x) + g(x)d(x) | (x3 + ax2 - 4x + 2) + (x3 + bx2 - 2)d(x) | 2x3 + (a + b) x2 - 4xd(x) | x 2x2 + (a + b) x - 4由，知a = 1，b = 3故(a，b) = (1，3)隨堂練習.與x3 + 有二次公因式，求a，b，p，q的值。【解答】a = 6，b = 7，p = 3，q = 2【詳解】設d(x) = x2 + px + q，則d(x) | (x3 + ax2 + 11x + 6) - (x3 + bx2 + 14x + 8)d(x) | (a - b) x2 - 3x - 2又d(x) | 4(x3 + ax2 + 11x + 6) - 3(x3 + bx2 + 14x + 8)d(x) | x x2 + (4a - 3b) x + 2x非已知多項式的因式d(x) | x2 + (4a - 3b) x + 2由，知a - b = - 1，4a - 3b = 3解，得a = 6，b = 7，而d(x) = x2 + 3x + 2p = 3，q = 29.兩多項式f (x)，g(x)的LCM為x4 + 4x3 - 15x2 - 58x - 40，f (x) + g(x) = 2x3 + 11x2 - x - 30，求f (x)及g(x)。【解答】f (x) = (x + 1)(x2 + 7x + 10)，g(x) = (x - 4)(x2 + 7x + 10)或f (x) = (x - 4)(x2 + 7x + 10)，g(x) = (x + 1)(x2 + 7x + 10)【詳解】設( f (x)，g(x) = d(x)且f (x) = d(x) h(x)，g(x) = d(x) k(x)，(h(x)，k(x) = 1則f (x) + g(x) = d(x) h(x) + k(x) = 2x3 + 11x2 - x - 30 f (x)，g(x) = d(x) h(x) k(x) = x4 + 4x3 - 15x2 - 58x - 40而d(x)為f (x) + g(x)與 f (x)，g(x)之HCF，由輾轉相除法得d(x) = x2 + 7x + 10h(x) + k(x) = 2x - 3，h(x) k(x) = x2 - 3x - 4 = (x + 1)(x - 4)故f (x) = (x + 1)(x2 + 7x + 10)，g(x) = (x - 4)(x2 + 7x + 10)或f (x) = (x - 4)(x2 + 7x + 10)，g(x) = (x + 1)(x2 + 7x + 10)隨堂練習.設f (x)，g(x)都是三次多項式且領導係數都是1，若其HCF與LCM的和為(x2 - 3x)2 + 4(x2 - 3x) + 3，求f (x)，g(x)。【解答】(x2 - 3x + 1)(x -1)，(x2 - 3x + 1)(x - 2)；(x2 - 3x + 3)(x)，(x2 - 3x + 3)(x - 3)（均可互調）【詳解】設HCF為d(x)，f (x) = d(x) m(x)，g(x) = d(x) n(x)，m(x)，n(x)互質HCF + LCM = d(x) + d(x) m(x) n(x) = (x2 - 3x)2 + 4(x2 - 3x) + 3d(x) 1 + m(x) n(x) = (x2 - 3x + 1)(x2 - 3x + 3)HCF | f (x)deg HCF deg f (x) = 3而deg (HCF + LCM) = 4deg LCM = 4deg d(x) = 2(i) d(x) = x2 - 3x + 1時1 + m(x) n(x) = x2 - 3x + 3m(x)n(x) = x2 - 3x + 2 = (x - 1)(x - 2)令m(x) = x -1，則n(x) = x - 2f (x) = (x2 - 3x + 1)(x -1)，g(x) = (x2 - 3x + 1)(x - 2)（可互調）(ii) d(x) = x2 - 3x + 3時1 + m(x) n(x) = x2 - 3x + 1m(x) n(x) = x2 - 3x = x(x - 3)令m(x) = x，則n(x) = x - 3f (x) = (x2 - 3x + 3)x，g(x) = (x2 - 3x + 3)(x - 3)（可互調）主題2：輾轉相除法1.輾轉相除法原理：若與為兩多項式，且，其中與為兩多項式，或，則。2.若，則，其中為任意多項式。重要範例1.設f (x) = x4 + x3 - 9x2 - 3x + 18，g (x) = x5 + 6x2 - 49x + 42，則f (x)與g (x)的HCF =。【解答】x2 + x - 6【詳解】由輾轉相除法得HCF為x2 + x - 62.若f (x) = g (x)Q (x) + r (x)，其中g (x) = 6x4 - 11x3 + 11x2 - 4x - 12，r (x) = 2x3 - x2 + 3x + 2，求f (x)與g (x)之最高公因式。 【解答】x2 - x + 2【詳解】由上式得(g (x)，r (x) = x2 - x + 2若f (x) = g (x)Q (x) + r (x)，由輾轉相除法原理得(f (x)，g (x) = (g (x)，r (x) = x2 - x + 23.設與的LCM為四次式，求實數a，b的值。【解答】a = - 7，b = - 4【詳解】令f (x) = x3 + x2 + ax + 2，g(x) = x3 + 2x2 + bx + 1，則deg f (x)g(x) = 6已知f (x)，g(x)之LCM為四次式，故其HCF為二次式由輾轉相除法得HCF為(a - b + 1) x2 + (a + 1) x + 2或x2 + (- a + b) x -1解得a = - 7，b = - 4隨堂練習. 設多項式d (x)為f (x) = 2x4 + x3 + 5x2 + x + 15與g (x) = 3x3 + 4x2 + 5x - 6之最高公因式，且已知其領導係數為1，求多項式d (x)。答：。 【解答】x2 + 2x + 3【詳解】由上輾轉相除法之計算，得最高公因式d (x) = x2 + 2x + 34.若二多項式與不互質，求k的值。 【解答】- 2，3【詳解】x2 - x + k與x3 + x2 + x + 3 + k不互質，設公因式d(x)，則de