Two questions on polynomial (F.4)

1.When a polynomial f(x) is divided by x-5,the remainder is 9.When it is divided by x+2,the remainder is -5.Find the remainder when f(x) is divided by (x-5)(x+2).
SOLUTION
let the remainder is ax+b
then
f(x)=g(x)(x-5)(x+2)+(ax+b)
Then by remainder theorem
f(5)=5a+b=9...(1)
f(-2)=-2a+b=-5...(2)
(1)-(2):
7a=14
a=2
sub into (2) b=2a-5=-1
So the remainder when f(x) is divided by (x-5)(x+2) is
2x-1
2.Let f(x) =3x^3 +mx^2 -nx -7 .When f(x) is divided by (x+1)(x-3),the remainder is 2x-4.

(a)Find the values of f(-1) and f(3).

(b)Set up two equations connecting m and n.

(c)Find the values of m and n.
SOLUTION
(a)
We have
f(x)=g(x)(x+1)(x-3)+2x-4
f(-1)=2(-1)-4=-6
f(3)=2(3)-4=2
(b)
Since f(-1)=-6
f(-1)=3(-1)^3 +m(-1)^2 -n(-1) -7=-6
-3+m+n-7=-6
m+n=4...(1)
This is the first equation
Since f(3)=2
f(3)=3(3)^3 +m(3)^2 -n(3) -7=2
81+9m-3n-7=2
9m-3n=-72
3m-n=-24
n-3m=24...(2)
This is the second equation
(c)
from (1)
m=4-n
sub into (2)
n-3(4-n)=24
-12+4n=24
n=9
m=4-9=-5

2007-04-18 00:33:48 補充：
答題中的 g(x) 就是 quotinet.應該用Q(x) 會好看些﹐不過其實都是同一樣東西。

1)
Assume the remainder be Ax + B where A and B are constants.
i.e. f(x) = Q (x-5)(x+2) + (Ax + B)
As given, f(5) = 9 and f(-2) = -5
f(5) = 5A + B = 9 --- (1)
f(-2) = -2A + B = -5 --- (2)

(1) - (2)
7A = 14
A = 2

Put it back into (1)
5(2) + B = 9
B = -1

So, remainder is 2x - 1
**********************
2)
a) Assume f(x) = Q(x+1)(x-3) + (2x-4) where Q is the quotient.
f(-1) = 2(-1) - 4 = -6
f(3) = 2(3) - 4 = 2

b) f(x) = 3x^3 + mx^2 - nx - 7
f(-1) = 3(-1)^3 + m(-1)^2 - n(-1) - 7 = -6
-3 + m + n - 7 = -6
m + n = 4 --- (1)

f(3) = 3(3)^3 + m(3)^2 - n(3) - 7 = 2
81 + 9m - 3n - 7 = 2
3m - n = -24 --- (2)

c)
(1) + (2)
4m = -20
m = -5
n = 9