想請問第12.題！我的解法完全不是答案想要的，想請問這樣解錯誤在哪裡呢？我先用有的條件假數出f(x)再假設乘了(x^2+1)的情況，餘式太高再除一次這樣。非常感謝🙏
N.已知實係數多項式f(x)的次數不超過兩次
k為正實數,又/(3) = 3k,則(9)= 52k
fiyy=k
f(x)=k
6
#f11)=K + L=K
f(x)= a(x-1)(x-¹) + b (x-1) + ©
(13)=3× ax=(x-1)× "fly=k-
K
b+k=k ⇒b=0
a (X-1) (X-1) + K = (1/
₁a+k=3K =) 2931
a=X
92
皆為化
【中科實中模擬考】
12. 設f(x)為三次實係數多項式,若滿足(1) = 3,f(2)=10,f(-1)=-17,f(4)=78求
【全國聯合模擬考】
(x²+1)f(x)除以x−3的餘式為
13. 實係數多項式 f(x)滿足f(1)=1,f(2) = 4,f(3)=9,f(4)=16,則下列敘述哪些是正
【中科實中模擬考】
確的?
(A)f(x)除以x-1的餘式為1
(B)對於任意實數x,都有f(x)=x²
(C)f(x)除以(x-2)(x-3)的餘式為x²
(D)令g(x)=f(x)-x²,則(x-1)(x-2)(x-3)(x-4)為g(x)的因式
(E)f(x)不可能為一個三次多項式
²² f(x) = a (x-1)(x-2)(x+1) + b(x-1)/(x-1) + ((x-1) + d
+ f(x) => x-1)(x-2)(x+1) - (x-1)(x-2) + 2(x-1) + 3₂
"
£13) =
Y
19
12
maid (C1)
Boitanidoo (H)
→ f (x) = (x-3) (8 (X) + $1
(X) (3)
31
STUJOUTE
of anotiziv bowolls roid
ali of zomerb
tupitib taom odiod of ti
-
D D G Y
borobianos (O)
Y
mesoniis ei sbivong won 29tole sousinovnos asol vise to
asgafonę gritoolios bus gaige
X (X²+1) 7 (x²+1) f(x) = (x-3) (@(x)) (X²+1) + 3) (X²+1) 9M sau o exat
ode doume dailgad bodil oda ydw bodes anW 542
gnir alanade stirw sonsbitnos may vod boqlad on
golondos Si bus agnining to
e boutest noindirxo
1-331+0+ 21
agaitnisq odi dow
liet t'abib3/93ibliod od oti bodesto tovinb aas
bus loodbe 93 +3ed nodw emexo diem list of boen
nummos si sxeuparus ada
of bamwulst (8)
boyonteab snow
26 bs
seorfw slqooq 101 totoda yoqats
not bolned (4)
****
43x+31)
#
no bobooqsb (3)
答應該是310吧