設實係數四次多項式其領導係數為1,如果F(1)=-

2011-05-09 7:36 pm
設F(x)為實係數四次多項式其領導係數為1,如果F(1)=-1、F(-2)=-4、F(3)=-9、
F(-4)=-16 求f(2)的值?

做好了多次,但算不出來 答案為-28,求過程

回答 (4)

2011-05-09 11:30 pm
✔ 最佳答案
顯然f(x)=-x^2會同時滿足f(1)=-1、f(-2)=-4、f(3)=-9、f(-4)=-16
但題目說f(x)是四次多項式,所以次數不夠
那麼便令f(x)=-x^2+a(x-1)(x+2)(x-3)(x+4)
又因領導係數為1,所以a=1
所以f(x)=-x^2+(x-1)(x+2)(x-3)(x+4)
f(2)=-4+1*4*(-1)*6=-28
2011-05-11 5:12 pm
請問克大師:
此題你怎麼想到f(x)=-x^2都滿足式子

若為f(-1)=1,f(2)=-8,f(-3)=27,f(4)=-64,其他條件同
它的關係式可否為
f(x)=-x^3+a(x+1)(x-2)(x+3)(x-4)
2011-05-09 9:30 pm
設F(x)為實係數四次多項式其領導係數為1,如果F(1)=-1、F(-2)=-4、
F(3)=-9、F(-4)=-16 求F(2)的值?
Sol
設 F(x)=(x-1)(x+2)(x-3)(x+4)+a(x-1)(x+2)(x-3)+b(x-1)(x+2)+c(x-1)-1
F(-2)=c(-2-1)-1=-4
c= 1
F (x)=(x-1)(x+2)(x-3)(x+4)+a(x-1)(x+2)(x-3)+b(x-1)(x+2)+(x-1)-1
=(x-1)(x+2)(x-3)(x+4)+a(x-1)(x+2)(x-3)+b(x-1)(x+2)+x-2
F(3)=b(3-1)(3+2)+3-2=-9
b=-1
F(x)=(x-1)(x+2)(x-3)(x+4)+a(x-1)(x+2)(x-3)-(x-1)(x+2)+x-2
=(x-1)(x+2)(x-3)(x+4)+a(x-1)(x+2)(x-3)-(x-1)(x+2)+x-2
=(x-1)(x+2)(x-3)(x+4)+a(x-1)(x+2)(x-3)-(x^2+x-2)+x-2
=(x-1)(x+2)(x-3)(x+4)+a(x-1)(x+2)(x-3)-x^ 2
F (-4)=a(-4-1)(-4+2)(-4-3)-16=-16
a= 0
F (x)=(x-1)(x+2)(x-3)(x+4)-x^ 2
F(2)=1*4*(-1)*6-4=-28


2011-05-09 8:47 pm
設f(x)為實係數四次多項式為x^4+ax^3+bx^2+cx+d
==>f(1)=1+a+b+c+d=-1
f(-2)=16-8a+4b-2c+d=-4
f(3)=81+27a+9b+3c+d=-9
f(-4)=256-64a+16b-4c+d=-16

==>a=2 ,b=-14,c=-14,d=24

f(x)=x^4+2x^3-14x^2-14x+24
==>f(2)=16+16-56-28+24=-28


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