關於pure math的問題...

1.
Let ax+b be the remainder when fx is divided by (x-2)(x-3).
Denote the quotient as Q.
Then, f(x) = (x-2)(x-3)Q + (ax+b)

The remainder is 5 when f(x) is divided by x-2, then f(2) = 5:
2a+b = 5 ...... (1)

The remainder is 9 when f(x) is divided by x-3, then f(3) = 9:
3a+b = 9 ...... (2)

(2)-(1):
a = 4

Substitute a=4 into (1):
2(4)+b = 5
b = -3

Ans: The required remainder is 4x-3

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2.
Since P(x) is divisible by x-1, then P(1) = 0.
(1)4+(a+10)(1)3+(10-b)(1)2+(1)+(16-c) = 0
1+a+10+10-b+1+16-c = 0
a-b-c = -38 ...... (1)

Q(x) is divisible by x2+x-2.
x2+x-2 = (x-1)(x+2)
Thus, Q(x) is divisible by both x-1 and x+2.

Since Q(x) is divisible by x-1, then Q(1) = 0.
(1)4+(b+10)(1)3+a(1)2+b(1)+c = 0
1+b+10+a+b+c=0
a+2b+c = -11 ...... (2)

Since Q(x) is divisible by x+2, then Q(-2) = 0.
(-2)4+(b+10)(-2)3+a(-2)2+b(-2)+c = 0
16-8b-80+4a-2b+c=0
4a-10b+c = 64 ...... (3)

(1)+(2):
2a+b = -49 ...... (4)

(1)+(3):
5a-11b = 26 ...... (5)

(4)x11+(5):
27a = -513
a = -19

Substitute a=-19 into (4):
2(-19)+b = -49
b = -11

Substitute a=-19 and b=-11 into (2):
(-19)+2(-11)+c = -11
c = 30

Ans: a=-19, b=-11, c=30
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