F.6 pure maths

回答在補充................

2010-09-03 11:15:02 補充：
1
Sol
f(x)=q(x)(x－1)(x－2)(x－3)+a(x－2)(x－3)+b(x－3)(x－1)+c(x－1)(x－2)
f(1)=a(－1)(－2)=2a
a=f(1)/2
f(2)=b(1)(1)=b
b=f(2)
f(3)=c(2)(1)=2c
c=f(3)/2

2010-09-03 11:15:44 補充：
1(b)
Sol
x^5+kx^2=p(x)(x－1)(x－2)(x－3)+mx+n
1+k=m+n
32+4k=2m+n
243+9k=3m+n
=>
(32+4k)－(1+k)=(2m+n)－(m+n)
31+3k=m
(243+9k)－(32+4k)=(3m+n)－(2m+n)
211+5k=m
211+5k=31+3k
2k=－180
k=－90

2010-09-03 11:16:07 補充：
2
Sol
f(x)+g(x)=(2x^4+x^3－6x^2－2x+3)+(2x^4－3x^3+2x－3)
=4x^4－2x^3－6x^2=2x^2(2x^2－x－3)
=2x^2(2x－3)(x+1)
f(x)－g(x)
=(2x^4+x^3－6x^2－2x+3)－(2x^4－3x^3+2x－3)
=4x^3－6x^2-4x+6
=2(2x^3－3x^2-2x+3)
=(x－1)(x+1)(2x－3)
So
the greatest common divisor of f(x) and g(x)=(x+1)(2x－3)

What's meant by "which the remainder when (x^5+kx^2) is divided by (x-1)(x-2)(x-3) constanst no term in x^2"??