最高公因式與最低公倍式匯總

1、§最高公因式典最低公倍式主jB 1 ：因式典倍式|1 .因式、倍式：f(x f(x)、g(x)懸二多 式,且 g(x)#0,若存在 q(x)使 f (x) = g(x) ,q(x),即耦g(x)懸f(x)之因式,f(x)懸g(x)之倍式,以g(x)f(x)。2 .公因式：若多式d(x)同畤是f(x)典g(x)之因式，U耦d(x)懸f(x)典g(x)之公因式。(雨 多式的公因式有多(0。)3 .互H：若多式f(x)典g(x)除常數(shù)多式外，沒(méi)有其他的公因式，JW耦它伸號(hào)互H。4 .最高公因式(H.C.F):段f (x) ,g(x)懸雨多式，若d(x)是它伸號(hào)的公因式中次數(shù)最高的，U d(

2、x)就？f (x), g(x)之最高公因式。常以H C ,F或(f (x) ,g(x)表小。(最高公因式不唯 一，但只差常數(shù)倍)5 .公倍式：鼓：f (x), g(x)都是非零多式,若m(x)同畤是f (x) , g(x)之倍式，m(x)懸 f (x) , g(x)之公倍式。6 .最低公倍式(L.C.M): f (x) ,g(x)的公倍式中,次數(shù)最小的，其最低公倍式,常以f(x), g(x)表示或以L C M表示。(最低公倍式不唯一，但只差常數(shù)倍)7 . f (x) =g(x)q(x) +r(x)= (f(x), g(x) = (g(x), r(x)8 .(f (x),g(x) =d(x),d

3、(x)m(x)f (x) n(x)g(x)9 .(f (x),g(x) f(x),g(x) = f (x) g(x)牛16定理【一次因式梅T，麻法】：f(x)=anxn+anxn十一 +a1x+a0懸整彳系數(shù)多式；a,b w Z,a =0,且(a,b) =1,若 axbS f ( x)之因式刖 aan,ba0 且 ab f (1) a + bf(1)。i重要靶例1 .若deg f (x) = 3, deg g (x)=3, T出下列正碓的敘述:(A) deg(f (x) g (x) = 6 (B)deg(f (x) g (x) = 3 (C) deg(f (x)g (x) = 6 (D) de

4、g(f (x)g (x) = 9(E)若f (x)典g (x)之最高公因式懸一次式，刖f (x)典g (x)之最低公倍式懸五次式®! deg f (x)表示多式f (x)的次數(shù)?！窘獯稹?C)(E)【群解】(A)(B)鰭。deg(f (x) + g (x) < 3(C)(D)deg (f (x)g(x) = degf (x) + deg g(x) = 3 + 3 = 6 .(C淺寸，(D)4g(E)封。由 f (x)g (x) = k (f (x), g (x)f (x), g (x)ndeg(f (x)g (x) = deg(f (x), g (x) + degf (x),

5、g (x) =6 = 1 + degf (x), g (x)故 degf (x), g (x) = 5隨堂維it忽deg f (x) = 5, deg g(x) = 3,刖下列敘述何者正碓？(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)【群解】(A) HCF | LCM . deg (HCF) < deg (LCM)(B) deg f (x)十 deg g(x) = deg

6、 (HCF) + deg (LCM) = 8(C)(D) deg (HCF)最大 =deg f (x), deg g(x)中之敕小者 u deg (LCM)最小(E) deg (LCM)最大 =deg f (x)十 deg g(x) = 82 .若 f (x) = 2x2 + ax + b, g(x) = 3x2 + cx + d 的 HCF x + 1, LCM 融 6x3 + 17x2 + 14x + e, a, b, c, d, e R, JW(A) a= 5 (B) b= 3 (C) c = 4 (D) d = 1 (E) e = 3。【解答】(A)(B)(C)(D)(E)【群解】1&

7、#176; HCF | LCM . x + 1 | 6x3 + 17x2 + 14x+e e= 32s LCM = 6x3 + 17x2+ 14x+ 3 = (x + 1)(2x + 3) (3x + 1)3s. 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 懸整數(shù)，f (x) =x3 -x2+ 4x + a - 7, g(x) = 2x3+ 5x2 - 7x + 2a - 6,已知 f (x)和 g(x)的最高公因式懸一次式，求a之

8、值?！窘獯稹?【群解】f (x) = x3 - x2 4x a - 7, g(x) = 2x3 5x2 - 7x 2a - 6ISf (x)典g(x)的最高公因式懸d(x)d(x) | g(x) - 2 f (x) = d(x) | 7x2 -15x 8 = d(x) | (7x - 8)(x - 1)d(x) = 7x - 8 或 x -1aZ d(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) = 0 1 -1 + 4 + a 7 = 0 a= 34 .段 a 懸整數(shù)，若 f (x) =x3+x2

9、 4x + (a- 7洌 g(x) = 2x3 7x2 + 7x + (2a- 8)的最高公因式懸一次式，JW a的值懸?！窘獯稹?【群解】已知 f (x) = x3 + x2 4x + (a- 7洌 g(x) = 2x3 7x2 + 7x + (2a 8)之 HCF 懸一次式，ig懸 d(x), JW d(x) | 2f (x) -g(x) = d(x) | 3 (3x2 - 5x - 2) = d(x) | 3(x - 2)(3x + 1)d(x) =x 一 2,或 d(x) = 3x 1(1)若 d(x) =x 2, JWx 2 | f (x)= f (2) = 0 =8 + 4 8 +

10、 (a - 7) = 0 = a = 31若數(shù)=雙十1，j3x十1 |的=”4)=0二 /+4+a_7=0 = a =包非整數(shù) 不合 故所求a之值懸3 27 9 3275 .段 kSI數(shù)，f (x) =x3 - 2x2 - 5x + 6, g(x) = x3 + k2x2 + 2kx16。(1) f (x)典g(x)有一次式的最高公因式畤，k之值懸何？(2) f (x)典g(x)有二次式的最高公因式畤，k之值懸何？【解答】k = - 5或-2k= 3 3232 2【群解】f (x) =x -2x -5x+6, g(x) = x +k x +2kx-16先符 f (x)因式分解得 f (x) =

11、 (x -1)(x + 2)(x - 3)x -1 | g(x)=g(1) = 0 = k 2+ 2k -15 = 0 = k = 3 或5x+ 2 | g(x) = g( - 2) = 0 3 4k 2 - 4k- 24 = 0 3 k = 3或-2x - 3 | g(x)= g(3) = 0 = 9k 2+ 6k + 11 = 0 =k 瓢數(shù)解(1) f (x)典g(x)有一次式的HCFk = 5 畤，HCFx1; k = - 2 畤，HCFx+ 2(2) k = 3 畤,HCF 二次式(x 1)(x+ 2)隨堂蜜里思 f (x) = x3 + ax2 - 5x + (a - 7)典 g(

12、x) = x2 + ax - 2 的 HCF 懸一次式，刖 a=。(a懸整數(shù))【解答】1【群解】IS f (x), g(x)的HCF懸d(x)a -7即J d(x) | f (x)-xg(x) = d- 3x (a - 7) =d- 3 (x-丁)_，一、2，一、a -7 (a -7) a(a -7)d(x) | g(x) = g(-) = 0 =1-r(o )-2-0393= 4a2 - 35a 31 = 0 =(a -1)(4a - 31) = 0a = 1 或 a =31,但 a整數(shù)a= 146 .若x2+px - 5典x3+px - 5的最大公因式懸一次式，求 p =o 【解答】4【群

13、解】d (x) | (x3 + px - 5)且 d (x) | (x2 + px - 5),刖 d (x) | (x3 + px - 5) 一(x2 + px - 5)即 d (x) | x2 (x 一 1), d (x) =x, x2均不合，取 d (x) =x -1(x - 1) | (x2 + px 一 5),即 1 +p - 5 = 0, p = 4隨堂糠1?.若多式f (x) =x2 + px + 6, g(x) =x3+px+ 6的LCM懸四次式，刖常數(shù)p之值懸?！窘獯稹?7【群解】1 ° f (x), g(x)之HCF懸一次式2 HCF | f (x), HCF |

14、g(x) . HCF | g(x) -f (x) =x3-x2=x2 (x-1)31但 f (0) = 6 豐 0 . HCF =x_14s 由f (1) = 0 p = 77 .已知二多邛(式x3 + ax2 + 2x +4典2x3 + x2 + a2x + 2有一(0二次公因式，卻J a的值懸 c【解答】2【群解】段 f (x) =x3+ax2+ 2x + 4, g(x) = 2x3 + x2 + a2x + 2已知f (x)典g(x)有一他I二次公因式，d(x), u d(x) | 2f (x) -g(x)= d(x) | (2a -1) x2 + (4 -a2) x + 6又 d(x)

15、 | f (x) - 2g(x) = d(x) | - x 3x2 (2 - a)x 2(a2 -1) 因 x非 f(x洌 g(x)的因式,所以 d(x) | 3x2 + (2 -a)x + 2(a2-1)2a -1 4 - a知=32 -a622(a2 -1)若2= 2,刖生二1=1, 6=3=1 a = 2懸其解32(a2 -1)32a -1 2 a 3右 a# 2, JW= = 2a -1 = 6 + 3a = a=- 731 a2 -12-7311r 人但 一 5 =豐2= a = - 7不合1(-7) -1 168 .鼓:f (x) = x3 + ax2 - 8x + 5, g (x

16、) = 2x3 + bx2 - 7x - 5,若 f (x)典 g (x)的最高公因式懸二次式, 求(1)哪慳寸(a, b) =。(2)最高公因式懸 。【解答】(1)(2, 7) (2) x2 + 3x - 5【群解】段 d (x) = (f (x), g (x),且 deg d (x) = 2川J 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 一 15由一d (x) = (2a - b)x2 - 9x 15由=d (x) = 3x2+(a +b

17、)x 15(丁 f (0)豐 0, g (0)豐 0)2a -b -9152a -b = -3. 二3a+b-15、a+b=9得數(shù)封(a, b) = (2, 7), HCF =d (x) =x2 + 3x 5暨邈里.雨多式f (x) =x3 + ax2 _ 4x + 2典g(x) =x3 + bx2 - 2的最高公因式懸二次式，刖數(shù)封(a, b) =?！窘獯稹?1, 3)【群解】IS f (x)典 g(x)的 HCF 融 d(x), JW d(x) | f (x) -g(x)二 d(x) | (x3 ax2 一 4x 2) - (x3 bx2 _ 2) = d(x) | (a - b) x2

18、- 4x 4d(x) | f (x) g(x) = d(x) | (x3 ax2 - 4x 2) (x3 bx2 - 2)二 d(x) | 2x3 + (a + b) x2 - 4x = d(x) | x 2x2 + (a + b) x - 4-44a b 4a-b=-2:=a = 1, b = 3a +b = 4故(a, b) = (1, 3)32322隨堂鐮iH? . x +ax +11x+6典 x + bx +14x+8 有二次公因式 x +px+q,求 a, b, p, q 的值?！窘獯稹縜 = 6, b = 7, p = 3, q = 2【群解】段d(x) =x2 px q,即d(x

19、) | (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 2v x非已知多式的因式.d(x) | x2+(4a- 3b) x + 2由 , 知-= = a-b=- 1 ,4a - 3b = 31 4a -3b 2解 ， 得2= 6, b = 7,而 d(x) =x2 + 3x + 2p = 3, q = 29.雨多式 f (x), g(x)的 LCM 懸 x4 + 4x3 15x2 58x

20、40, f (x) + g(x) = 2x3 + 11x2 - x - 30,求 f (x)及 g(x)?！窘獯稹縡 (x) = (x + 1)(x2 + 7x + 10), g(x) = (x - 4)(x2 + 7x + 10)或 f (x) = (x - 4)(x2 + 7x + 10), g(x) = (x1)(x2 7x 10)【群解】意：(f (x), g(x) = d(x)且 f (x) = d(x) h(x), g(x) = d(x) k(x), (h(x), k(x) = 1川J f (x) g(x)=d(x) h(x) + k(x)= 2x311x2 - x - 30f (

21、x), g(x) = d(x) h(x) k(x) = x4 + 4x3 - 15x2 - 58x - 40而 d(x)懸 f (x) + g(x)則f (x), g(x)之HCF,由Wl相除法得d(x)=x2 7x 10=h(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)f (x), g(x)都是三次多式且加翼系數(shù)都是1

22、，若其HCF典LCM的和懸(x2 3x)22+ 4(x 3x) + 3,求 f (x), g(x)0【解答】(x2 - 3x+ 1)(x -1), (x2 - 3x + 1)(x- 2); (x2 - 3x+ 3)(x), (x2- 3x + 3)(x- 3)(均可互加【群解】鼓:HCF 融 d(x), f (x) = d(x) m(x), g(x) = d(x) n(x), m(x), n(x)互 HCF LCM = d(x) d(x) m(x) n(x) = (x2 - 3x)2 4(x2 - 3x) 3 d(x) 1 m(x) n(x) = (x2 - 3x 1)(x2 - 3x 3)v

23、 HCF | f (x) . deg HCF < deg f (x) = 3而 deg (HCF + LCM) = 4 . deg LCM = 4 = deg d(x) = 2(i) d(x) = x2 - 3x 1 畤1 +m(x) n(x) = x 3 3x + 3 . m(x)n(x) =x2 - 3x + 2 = (x - 1)(x - 2)令 m(x)= x -1,刖 n(x)= x - 2= f (x) = (x2 - 3x + 1)(x -1), g(x) = (x2 - 3x + 1)(x - 2)(可互湖)(ii) d(x) =x2- 3x + 3 畤1 +m(x) n

24、(x) = x2 3x + 1 m(x) n(x) = x2 3x = x(x - 3)令 m(x) =x,刖 n(x) =x 3= f (x) =(x2 3x+ 3)x, g(x) = (x2 - 3x+ 3)(x- 3)(可互湖)9主jB 2：樽相除法1 .H樽相除法原理：若f(x)典g(x)懸雨多式,且f(x) =g(x)q(x) + r(x),其中q(x)典r(x)懸 麗多項(xiàng)式,r(x) =0或degr(x) <degg(x),即"f (x), g(x) = (g(x), r(x)。2 .若 d(x) f (x) , d(x) g(x),即J d(x) m(x) f (

25、x) ±n(x)g(x),其中 m(x), n(x)懸任意多式。重要靶例I； (x) =x4 + x3 9x2 3x + 1& g (x) =x5 + 6x2 49x + 42 即J f (x)典 g (x)的 HCF =【解答】x2 x - 6【群解】1 + 1-9-3+181+ 0 7 + 61 + 0+ 0 + 6 49 + 421 十 1 9-3 + 181-2-9+181+07+ 6-1 + 9 + 9-67 + 42-1 - 1+9+ 3- 18-2-2-2+1210 10 + 0 70+ 601 + 1 61十0 7十61 + 161 - 1 + 6 1 + 6

26、0I 1由Wl相除法得HCF懸x2+x- 62.若 f (x) = g (x), Q (x) + r (x),其中 g (x) = 6x4 - 11x3 + 11x2 - 4x 一 12, r (x) = 2x3 - x2 + 3x 十 2, 求f (x)典g (x)之最高公因式 ?！窘獯稹縳2-x + 2【群解】3-46-11+11- 4-12 2-1-3 + 26 3+ 9+ 62-2 + 42+ I8 + 2 10-12-8+ 4-12- 82| - 2 + 2 - 4 1-1+21-1 + 2I - 1 + 20由上式得(g (x), r (x) =x2 -x + 2若 f (x) =

27、g (x)Q (x)+r(x),由H樽相除法原理得(f (x), g (x) = (g (x), r (x) = x2 - x+ 23.1£x3+x2 +ax+2 典 x3+2x2+bx+1 的 LCM 懸四次式，求數(shù) a, b 的值?！窘獯稹縜 = - 7, b = - 4【群解】令 f (x) =x3 +x2 + ax + 2, g(x) = x3 + 2x2 + bx + 1,刖 deg f (x). g(x) = 6 已知f (x), g(x)之LCM懸四次式，故其 HCF 二次式由®I樽相除法得 HCF 雙a 一 b + 1) x2 + (a + 1) x + 2 或 x2 + (- a + b) x -1I 1+1-21 B +