中四 polynomials

Since the degree of the remainder must be less than that of divisor , so if the degree of the divisor is 2 , then the possible maximum degree of the remainder is 1 ,hence we let the remainder is (ax+b) but not a constant.
When a polynomial p(x) is divided by x-1 and x+1, the remainders are 1 and 3 respectively
p(x)=(x-1)Q1(x)+1---(1)
p(x)=(x+1)Q2(x)+3---(2)
put x=1 into (1) ,p(1)=1
put x=-1 into (2) ,p(-1)=3
And now , let Q(x) and ax + b be the quotient and the remainder respectively when p(x) is divided by x^2 -1
p(x)=(x^2-1)Q(x)+(ax+b)---(3)
put x=1 into (3),
a+b=p(1)=1---(4)
put x=-1 into (3),
-a+b=p(-1)=3---(5)
(4)+(5):
2b=4
b=2
put b=2 into (4),
a=-1
Therefore , the remainder is (-x+2)

method 1
set p(x)=q(x)(x-1)(x+1)+ax+b
so p(1)=a+b=1
p(-1)=-a+b=3
so (a+b)+(-a+b)=1+3
so b=2
so a=-1
the remainders are -x+2

method 2
set p(x)=q(x)(x-1)(x+1)+a(x-1)+b
p(1)=b=1
p(x)=q(x)(x-1)(x+1)+a(x-1)+1
p(-1)=a(-2)+1=3
so a=-1
so

1. p(x) = (x-1)q(x) + 1
p(x) = (x+1)Q(x) + 3


p(1) = 1
p(-1) = 3


Now, suppose the remainder is Ax+B when p(x) is divided by
(x-1)(x+1).


p(x) = (x-1)(x+1)q(x) + Ax + B.


put p(1) = 1

1 = A+B (1)

put p(-1) = 3

3 = -A+B (2)


(2)-(1)

2 = -2A

A = -1

put A = -1 (1)

B = 2

so, Ax+B = -x+2


2. 一般情況應該都會用 ax+b,