Polynomials' problems

1. (a) f(x) = x^3 + ax^2 + bx + 12
f(x) is divisible by x^2+x – 6
Let f(x) = Q(x) ( x^2 + x – 6 ) , where Q(x) is a polynomial in x
f(x) = Q(x) (x – 2 ) ( x + 3 )
f(2) = Q(2) (2 – 2 ) ( 2 + 3 ) = 0
f(2) = 2^3 + a x (2)^2 + b x (2) + 12 = 0
8 + 4a + 2b + 12 = 0
2a + b + 10 = 0 . . . . . . . .(1)
f(–3) = Q(–3) (–3 – 2 ) (–3 + 3 ) = 0
f(–3) = (–3)^3 + a x (–3)^2 + b x (–3) + 12 = 0
–27 + 9a – 3b + 12 = 0
3a – b – 5 = 0 . . . . . . . . (2)
(1) + (2) : 5a + 5 = 0
5a = –5
a = –1
b = –8

(b) f(x) = x^3 – x^2 – 8x + 12 = ( x – 2 )^2 ( x + 3 )
f(x–1) = [ ( x – 1 ) – 2 ]^2 [ ( x – 1 ) + 3 ]
= ( x – 3 )^2 ( x + 2 )
[f(x–1)]^2 = ( x – 3 )^4 ( x + 2 )^2
Let g(x) = [f(x–1)]^2 = ( x – 3 )^4 ( x + 2 )^2
when [f(x–1)]^2 , i.e. g(x), is divided by x – 1, the remainder is :
g(1) = ( 1 – 3 )^4 x ( 1 + 2 )^2 = (–2 )^4 x 3^2 = 16 x 9 = 144

2. (b) Function g(x) is missing.



2009-07-20 02:03:27 補充：
2. (a) f(x) = x^3 + ax^2 – 5x + b
f(–1) = 0
(–1)^3 + a (–1)^2– 5 (–1) + b = 0
(–1) + a + 5 + b = 0
a + b + 4 = 0 . . . . . (1)
f(2) = –9
2^3 + a (2)^2– 5 (2) + b = –9
8 + 4a – 10 + b = –9
4a + b + 7 = 0 . . . . (2)
(2) – (1) : 3a + 3 = 0
3a = –3
a = –1
b = –3

2009-07-20 02:28:29 補充：
(b1) From (a), f(x) = x^3 – x^2 – 5x – 3 = ( x + 1 )^2 ( x – 3 )
g(x) = (x – 3 ) f(x)
= ( x + 1 )^2 ( x – 3 )^2
g(–1) = (–1 + 1 )^2 ( –1 – 3 )^2 = 0
g(2) = ( 2 + 1 )^2 ( 2 – 3 )^2 = ( 3^2 ) ( –1 )^2 = 9

2009-07-20 02:30:16 補充：
(b2) g(x) = ( x + 1 )^2 ( x – 3 )^2
x^2 – x – 2 = ( x – 2 ) ( x + 1 )
When g(x) is divided by x^2 – x – 2, the remainder is cx + d
Let g(x) = P(x) ( x^2 – x – 2 ) + cx + d
g(x) = P(x) ( x – 2 ) ( x + 1 ) + cx + d
g(2) = P(2) ( 2 – 2 ) ( 2 + 1 ) + 2c + d = 2c + d

2009-07-20 02:30:33 補充：
From (a), g(2) = 9
2c + d = 9 . . . . . . (1)
g(–1) = P(–1) (–1 – 2 ) (–1 + 1 ) – c + d = d – c
From (a), g(–1) = 0
d – c = 0 . . . . . . . (2)
(1) – (2) : 3c = 9
c = 3
d = 3

2009-07-20 02:34:00 補充：
I am sorry that I make a mistake in the last part, it should be:

From (b1), g(2) = 9
2c + d = 9 . . . . . . (1)
g(–1) = P(–1) (–1 – 2 ) (–1 + 1 ) – c + d = d – c
From (b1), g(–1) = 0
d – c = 0 . . . . . . . (2)
(1) – (2) : 3c = 9
c = 3
d = 3

Reply for part (b2).
From part (b1) we know g(-1) = 0 and g(2) = 9.
When g(x) is divided by x^2 - x - 2 = (x - 2)(x + 1), the expression will be
g(x) = (x -2)(x + 1)Q(x) + cx + d where Q(x) is the quotient.
Put x = 2
g(2) = (2 - 2)(2 + 1) Q(2) + 2c + d = 2c + d = 9...................(1)
Put x = -1
g(-1) = (-1 - 2)(-1 + 1)Q(-1) -c + d = - c + d = 0....................(2)
(1) - (2) we get
3c = 9, so c = 3.
From (2) d = c = 3.

(b)Let g(x)=(x-3)f(x). When g(x) is divided by x^2-x-2, the remainder is cx+d.

thx~