F.4 MATH !!!!!! 急 20點=]

Chp.7 PEQ 8
as f(x) = (x^3 - 7x + 6) is divisible by (x^2 - 3x + k)
let f(x) = (x + a) (x^2 - 3x + k) where a is integer

(x + a) (x^2 - 3x + k) = (x^3 - 7x + 6)
x^3 - 3x^2 + kx + ax^2 - 3ax + ak = x^3 - 7x + 6
x^3 + (a - 3) x^2 + (k - 3a)x + ak = x^3 - 7x + 6
compare the coefficient of similar terms
a - 3 = 0
a = 3

k - 3a = -7
k = -7 + 3(3) = -7 + 9 = 2

ak = 6(it is for checking)

so, k=2

Chp.7 suppex.
1. A (x + 1) (2x + 1) (x - 2) + C (x - 2)(x - 1) = x^2 + 2x - 1
there is no B, probably there is an error in the question. please check again.

2. A(x + 1) (x - 3) + B (4x - 1) + 5 = 5x^2 - 6x + C

*method 1(comparsion)

Ax^2 - 2Ax - 3A + 4Bx - B + 5 = 5x^2 - 6x + C
Ax^2 + (4B - 2A)x + (-3A - B + 5) = 5x^2 - 6x + C

compare the coefficient of similar terms
A = 5
4B - 2A = -6
4B - 2(5) = -6
4B - 10 = -6
B = 1
-3A - B + 5 = C
C = -3(5) - 1 + 5 = -11

so, A = 5, B = 1, C = -11

method 2(substitution)

(for your reference, for example, to eliminate A
choose x = -1 or 3 so that x+1 = 0 or x - 3 = 0)

when x = -1,
-5B + 5 = 5 + 6 + C
-5B = 6 + C
5B + C = -6 (equation 1)
when x = 3
11B + 5 = 5(3)^2 - 6(3) + C
11B - C = 22 (equation 2)
when x = 1/4,
A (1/4 + 1) (1/4 - 3) + 5 = 5(1/4)^2 - 6(1/4) + C
(-55/16) A + 5 = 5/16 - 3/2 +C
55A + 16C = 99 (equation 3)

equation 1 + equation 2:
16B = 16
B = 1
substitute B = 1 into equation 1,
5(1) + C = -6
C = -11
substitute C = -11 into equation 3,
55A + 16(-11) = 99
A = 5

so, A = 5, B = 1, C = -11

*you can see that method 2 is so complicated, while method 1 is simple.
method 2 is recommended.
(sometimes, method 2 is better when you see Chp.7 suppex. Q3)

2013-06-05 09:18:38 補充：
3. x^3 = p (x+1) (x+2) (x+3) + q (x+2) (x+3) + r (x+3) + s

method 2 in Q2

when x = -3,
s = -27
when x = -2,
(-2)^3 = r (-2+3) - 27
r = 19
when x = -1,
(-1)^3 = q (-1+2) (-1+3) + r (-1+3) + s
-1 = 2q + 2r + s
2q = -1 -2(19) - (-27)
q = -6
when x = 0,
0 = 6p +6q +3r + s
p = 1