f.4 maths ( 10 PTS )

1 a) show that x+2 is a factor of (x+1)^(2n+1)+1 , where n is a positive integer.
b) If n is a positive integer , determine whether 6^(2n +1 ) is divisible by 7. explain your assertion.
ANSWER
a) Show that x+2 is a factor of (x+1)^2n+1 +1 where n is a positive integer .

f(x) = (x+1)^(2n+1) + 1
f(-2) = (-2+1)^(2n+1) + 1 = -1^(2n+1) + 1
since 2n+1 is odd number (n is a positive integer),
therefore -1^(2n+1) = -1
--> -1+1=0
By factor theorem therefore x+2 is a factor of (x+1)^(2n+1) +1.

b) If n is a positive integer, determine whether 6^2n+1 +1 is divisible by 7
Explain your assertion
Method 1
substitute x=5 in (x+1)^(2n+1)+1 , then (x+1)^(2n+1)+1=6^(2n +1 ) +1 which is divisible by (x+2)=7 by the result of (a)
Method 2
Using mathematical induction
when n=1
6^(2n+1) + 1 = 217 which is divisible by 7

assume n=k is true
6^(2k+1) + 1 which is divisible by 7

by induction, when n=k+1
6^(2(k+1)+1) + 1 = 6^(2k+3) + 1 which is divisible by 7
2) let P(x) be a polynomial . when P(x) is divided by (x-1 ) , the remainder is 5. when P(x) is divided by (x+2) , he remainder is -1 . if P(x) id divided by (x^2 + x +2) , the remainder is Ax+B , find the val of A and B .
ANSWER
When P(x) is divided by x-1, the remainder is 5.
Therefore, by Remainder Theorem, P(1) = 5

When P(x) is divided by x+2, the remainder is -1
Therefore, by Remainder Theorem, P(-2) = -1

By Division Algorithm (除法), we can find Q(x) (商) and R(x) (餘數) such that
P(x) = Q(x) (x²+x-2) + R(x) -------------(*)
where deg R(x) ＜ deg(x²+x-2) = 2
From the question, we know that R(x) = Ax+B

Substitue x = 1 into (*)
P(1) = Q(1)[(1)²+(1)-2] + A(1) + B
5 = 0 + A + B
A+B = 5 -----------------(1)

Substitute x = -2 into (*)
P(-2) = Q(-2)[(-2)²+(-2)-2] + A(-2) + B
-1 = 0 - 2A + B
2A - B = 1 -----------------(2)

(1) + (2):
3A = 6
A = 2

Substitute into (1):
2+B=5
B = 3