f.4 maths

a.
∵f(x)=3x^3+mx^2-nx-7
f(x)=Q(x)．(x+1)(x-3)+(2x-4)
f(-1)=-6
f(3)=2

b.
∴3(-1)^3+m(-1)^2-n(-1)-7=-6
-3+m+n=-6
m+n=-3
3(3)^3+m(3)^2-n(3)-7=2
81+9m-3n-7=2
3m-n=-24

c.
From b),
m+n=-3------------------1
3m-n=-24----------------2
From 1), m=-3-n------3
Sub 3 into 2, 3(-3-n)-n=-24
-9-3n-n=-24
-4n=-15
n=15/4-------------------4
Sub 4 into 1, m+15/4=-3
m=-27/4

2007-03-29 00:06:33 補充：
Calculation mistakes= =b.∴3(-1)^3 m(-1)^2-n(-1)-7=-6-3 m n-7=-6m n=43(3)^3 m(3)^2-n(3)-7=281 9m-3n-7=23m-n=-24

2007-03-29 00:06:46 補充：
c.From b), m n=4-------------------13m-n=-24----------------2From 1), m=4-n------3Sub 3 into 2, 3(4-n)-n=-2412-3n-n=-24-4n=-36n=9-------------------4Sub 4 into 1, m 9=4m=-5