高一數學(多項式)

1.為何x^2-1是x^18-1的因式?
因x^2-1=(x+1)(x-1)
令f(x)=x^18-1
==>f(1)=1-1=0 ,f(-1)=1-1=0 餘式定理

所以x^2-1是x^18-1的因式

2015-08-02 20:51:33 補充：
4.f(x-1)-f(x-2)=2x-3
f(0)=2
求f(x)=?
令f(x)=ax^2+bx+c
==>f(x-1)=a(x-1)^2+b(x-1)+c ,f(x-2)=a(x-2)^2+b(x-2)+c

f(x-1)-f(x-2)=2x-3
==>2ax-3a+b=2x-3
==>a=1,b=0

f(x)=x^2+c
==>f(0)=0+c=2 ==>c=2

所以f(x)=x^2+2

2015-08-02 21:18:23 補充：
3.f(x)為完全平方式
f(x)=x^4+4x^3-2x^2+ax+b
g(x)=x^2+cx+d
且g(x) | f(x)
求c+d=?
令f(x)=(x^2+ex+f)^2=x^4+2ex^3+(e^2+2f)x^2+2ef+f^2=x^4+4x^3-2x^2+ax+b
==>e=2,f=-3 ,a=-12,b=9

f(x)=(x^2+2x-3)^2
因g(x) | f(x) ==>g(x)=x^2+2x-3=x^2+cx+d ==>c=2,d=-3

所以c+d=2-3=-1

2015-08-02 21:37:13 補充：
2.a,b,c為整數

f(x)=(x-a)(x-10)+1
g(x)=(x-b)(x-c)

若f(100)=g(100)
f(101)=g(101)
f(102)=g(102)

求a+b+c=?
因代入值三點以上都相等表同函數
f(x)=g(x)
==>(x-a)(x-10)+1=(x-b)(x-c)
==>x^2-(a+10)x+(10a+1)=x^2-(b+c)x+bc
==>a+10=b+c, bc=10a+1
==>(10-c)(b-10)=-1
==>(a,b,c)=(8,9,9)或(12,11,11)
a+b+c=26或34

2015-08-02 21:41:14 補充：
1.
因x^2-1=(x+1)(x-1)
令f(x)=x^18-1
==>f(1)=1-1=0 ,f(-1)=1-1=0 餘式定理所以x^2-1是x^18-1的因式
2.
因代入值三點以上都相等表同函數
f(x)=g(x)
==>(x-a)(x-10)+1=(x-b)(x-c)
==>x^2-(a+10)x+(10a+1)=x^2-(b+c)x+bc
==>a+10=b+c, bc=10a+1
==>(10-c)(b-10)=-1
==>(a,b,c)=(8,9,9)或(12,11,11)
a+b+c=26或34

2015-08-02 21:43:53 補充：
3.
令f(x)=(x^2+ex+f)^2=x^4+2ex^3+(e^2+2f)x^2+2ef+f^2=x^4+4x^3-2x^2+ax+b
==>e=2,f=-3 ,a=-12,b=9
f(x)=(x^2+2x-3)^2
因g(x) | f(x) ==>g(x)=x^2+2x-3=x^2+cx+d ==>c=2,d=-3
所以c+d=2-3=-1

2015-08-02 21:44:12 補充：
4.
令f(x)=ax^2+bx+c
==>f(x-1)=a(x-1)^2+b(x-1)+c ,f(x-2)=a(x-2)^2+b(x-2)+c
f(x-1)-f(x-2)=2x-3
==>2ax-3a+b=2x-3
==>a=1,b=0
f(x)=x^2+c
==>f(0)=0+c=2 ==>c=2
所以f(x)=x^2+2

2015-08-02 22:11:23 補充：
感謝Lopez大師指正

3.更正

f = (x^2+2x-3)^2 = (x-1)^2(x+3)^2
g(x) 有三種可能:
g(x) = (x-1)^2 = x^2-2x+1 , c+d = -1
g(x) = (x+3)^2 = x^2+6x+9 , c+d = 15
g(x) = (x-1)(x+3) = x^2+2x-3 , c+d = -1

Ans: c+d = 15 或 -1

2015-08-02 22:35:10 補充：
a+10=b+c,bc=10a+1 代入削去法
==>10a+100=10b+10c,10a=bc-1
==>bc-1+100=10b+10c
==>10b+10c-bc=99
==>b(10-c)-10(10-c)+100=99 湊提出公因數
==>(10-c)(b-10)=-1










會變成 (10-c)(b-10)=-1

3.
f
= (x^2+px+q)^2
= x^4+2px^3+(p^2+2q)x^2+2pqx+q^2
= x^4+4x^3-2x^2+ax+b

比較係數得 p = 2 , q = - 3
f = (x^2+2x-3)^2 = (x-1)^2(x+3)^2

g 有三種可能:
g = (x-1)^2 = x^2-2x+1 , c+d = -1
g = (x+3)^2 = x^2+6x+9 , c+d = 15
g = (x-1)(x+3) = x^2+2x-3 , c+d = -1
Ans: c+d = 15 或 -1

第２題：

兩個二次函數若能夠有兩點以上的相交點，即代表兩個二次函數是恆等的！

解決了！