maths chap7 f4 15points

1)
Let f(x)=x^3+(a+4)x^2-2x-1

When x^3+(a+4)x^2-2x-1 is divided by x-1,
the remainder is f(1)=1+(a+4)-2-1=a+2

When x^3+(a+4)x^2-2x-1 is divided by x+a,
the remainder is f(-a)=(-a)^3+(a+4)(-a)^2-2(-a)-1=-a^3+a^3+4a^2+2a-1=4a^2+2a-1

Since the remainders are equal, then:
4a^2+2a-1=a+2
4a^2+a-3=0
a=3/4 or a=-1

2)
Since when P(x) is divided by x-1 and x-2, the remainders are -4 and -28 respectively, we have P(1)=-4 and P(2)=-28

We then let the remainder when P(x) is divided by (x-1)(x-2) be (ax+b),
and P(x)=(x-1)(x-2)Q(x)+(ax+b), for some polynomial Q(x)

Sub. x=1 and x=2 into P(x) respectively, we have:
P(1)=-4=a+b
P(2)=-28=2a+b

Solving, we have a=-24 and b=20, then the remainder is (-24x+20)

3)
From the given information, we have x^99+2=(x+1)Q(x)+1, where Q(x) is quotient

Sub. x=9 into x^99+2=(x+1)Q(x)+1:
9^99+2=10Q(x)+1
9^99=10Q(x)-1

Hence, the remainder is 10-1=9


2011-01-21 00:11:43 補充：
For the last part, it should be written as:

Sub. x=9 into x^99+2=(x+1)Q(x)+1:
9^99+2=10Q(9)+1
9^99=10Q(9)-1
9^99=10Q(9)-10+9
9^99=10[Q(9)-1]+9

Hence, the remainder is 9