急~急~急~數學恆等多項式

Q.1
P(x-1)(x-2)+Q(x-2)(x+1)+R(x+1)(x-1)≡4x²-8x+6
P(x²-3x+2)+Q(x²-x-2)+R(x²-1)≡4x²-8x+6
(P+Q+R)x²+(-3P-Q)x+(2P-2Q-R)≡4x²-8x+6
So,　 P+Q+R = 4 ................. (a)
　　　-3P-Q = -8 .................. (b)
　　　2P-2Q-R = 6 ............... (c)
(a)+(c): 3P-Q = 10 ................ (d)
(d)-(b): 6P = 18, i.e. P = 3.
Put into (d): Q = 3P-10 = -1.
Put into (a): R = 4-P-Q = 2.

2007-02-01 21:20:46 補充：
Q.2A(x 1)² B(x 1)(x-2) C(x-2)²≡15x-12(A B C)x² (2A-B-4C)x (A-2B-4C)≡15x-12So,　 A B C = 0 ................. (a)　　　2A-B-4C = 15 .............. (b)　　　A-2B-4C = -12 ............. (c)

2007-02-01 21:20:56 補充：
4(a) (b): 6A 3B = 15　　　　2A B = 5 ................ (d)4(a) (c): 5A 2B = -12 ........... (e)(e)-2(d): A = -22.Put into (d): B = 5-2A = 49.Put into (a): C = -A-B = -27.

2007-02-01 21:23:50 補充：
It seems there's some problem with displaying the ＋ sign.The missing signs in my Q.2 answer are ＋.

2007-02-01 21:31:15 補充：
Q.3(a):P(x＋1)(x＋2)＋Q(x＋2)(x＋3)＋R(x＋3)(x＋4)≡x²＋8x＋14(P＋Q＋R)x²＋(3P＋5Q＋7R)x＋(2P＋6Q＋12R)≡x²＋8x＋14So,　 P＋Q＋R = 1 ................. (a)　　　3P＋5Q＋7R = 8 ............ (b)　　　2P＋6Q＋12R = 14　　　i.e. P＋3Q＋6R = 7 ........ (c)

2007-02-01 21:31:33 補充：
6(a)-(c): 5P＋3Q = -1 ............... (d)7(a)-(c): 4P＋2Q = -1 ............... (e)(d)-(e): P＋Q = 0 ...................... (f)(e)-2(f): 2P = -1　　　 P = -0.5Put into (f): Q = -P = 0.5Put into (a): R = 1-P-Q = 1

2007-02-01 21:36:40 補充：
Q.3(b)To solve: -2(x＋1)(x＋2)＋2(x＋2)(x＋3)＋4(x＋3)(x＋4)＝3x²＋20x＋29.Solution:From Q.3(a),　-0.5(x＋1)(x＋2)＋0.5(x＋2)(x＋3)＋1(x＋3)(x＋4)≡x²＋8x＋14　i.e. -2(x＋1)(x＋2)＋2(x＋2)(x＋3)＋4(x＋3)(x＋4)≡4x²＋32x＋56.

2007-02-01 21:36:55 補充：
Now,　-2(x＋1)(x＋2)＋2(x＋2)(x＋3)＋4(x＋3)(x＋4)＝3x²＋20x＋29　i.e. 4x²＋32x＋56 ＝ 3x²＋20x＋29　i.e. x²＋12x＋27 ＝ 0　(x＋3)(x＋9) ＝ 0　x = -3 or x = -9.

2007-02-01 21:38:31 補充：
這裡有太多題目了，我覺得應該分開３個 threads 來問，這樣會更快得到正確答案。

P(x-1)(x-2)+Q(x-2)(x+1)+R(x+1)(x-1)≡4x^2 - 8x+6
Put x=2,
L.H.S.=P(2-1)(2-2)+Q(2-2)(2+1)+R(2+1)(2-1)=3R
R.H.S.=4*2^2 - 8*2+6=16-16+6=6
R=2
Put x =1,
L.H.S.=P(1-1)(1-2)+Q(1-2)(1+1)+R(1+1)(1-1)=-2Q
R.H.S.=4*1^2 - 8*1+6=4-8+6=2
Q=-1
Put x =-1,
L.H.S.=P(-1-1)(-1-2)+Q(-1-2)(-1+1)+R(-1+1)(-1-1)=3P
R.H.S.=4*(-1)^2 - 8*(-1)+6=4+8+6=18
P=6


A(x+1)^2 + B(x+1)(x-2)+C(x-2)^2≡15x-12
Put x=-1,
L.H.S.=A(-1+1)^2 + B(-1+1)(-1-2)+C(-1-2)^2=9C
R.H.S.=15(-1)-12=-15-12=-27
C=-3
Put x=2,
L.H.S.=A(2+1)^2 + B(2+1)(2-2)+C(2-2)^2=9A
R.H.S.=15*2-12=30-12=18
A=2
Put x=0,
L.H.S.=A(0+1)^2 + B(0+1)(0-2)+C(0-2)^2=A-2B+4C=2-2B-12=-2B-10
R.H.S.=15*0-12=-12
B=1


P(x+1)(x+2)+Q(x+2)(x+3)+R(x+3)(x+4)≡x^2+8x+14
Put x=-3,
L.H.S.=P(-3+1)(-3+2)+Q(-3+2)(-3+3)+R(-3+3)(-3+4)=2P
R.H.S.=(-3)^2+8(-3)+14=9-24+14=-1
P=-0.5
Put x=-2,
L.H.S.=P(-2+1)(-2+2)+Q(-2+2)(-2+3)+R(-2+3)(-2+4)=2R
R.H.S.=(-2)^2+8(-2)+14=4-16+14=2
R=1
Put x=-1,
L.H.S.=P(-1+1)(-1+2)+Q(-1+2)(-1+3)+R(-1+3)(-1+4)=2Q+6R=2Q+6
R.H.S.=(-1))^2+8(-1)+14=1-8+14=7
Q=0.5


(b)-2(x+1)(x+2)+2(x+2)(x+3)+4(x+3)(x+4)=3x^2+20x+29
When we multiply (A) by 4,we have
-2(x+1)(x+2)+2(x+2)(x+3)+4(x+3)(x+4)=4x^2+32x+56
Therefore,3x^2+20x+29=4x^2+32x+56
-x^2-12x=-27
x=-3

Put x = -1, we have
P(-1 - 1)(-1 - 2) ≡ 4 + 8 + 6
i.e. P = 3

Put x = 1, we have
Q(1 + 1)(1 - 2) ≡ 4 - 8 + 6
i.e Q = -1

Put x = 2, we have
R(2 + 1)(2 - 1) ≡ 4 * 2^2 - 8^2 + 6
R = 3

(N.B.: sometimes, putting some particular values of x into the identity to find unknowns constant will be much easier than expand the whole expansion.)

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So, for the 2nd questions, put x = -1 and 2 respectively, we get
C (-1 - 2)^2 = 15(- 1) - 12 and
A(2 + 1)^2 = 15(2) - 12
i.e. A = 2, C = - 3
Put x = 0, we have
A - 2B + 4C = -12, B = 1

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(A) Put x = -2, -3 respectively, we get
R(-2 + 3)(-2 + 4)≡4 - 16+14
P(-3 + 1)(-3 + 2)≡9 - 24 + 14
So solving we have
R = 1, P = -1/2

Hence, put x = 0, we have
2P + 6Q + 12R = 14, Q = 1/2

(B), using (A), we have
-2(x+1)(x+2)+2(x+2)(x+3)+4(x+3)(x+4)=3x^2+20x+29
4 * [-1/2 * (x+1)(x+2)+1/2 * (x+2)(x+3) + (x+3)(x+4)]=3x^2+20x+29
4 *(x^2 + 8x + 14) = 3x^2 + 20x + 29 [Using (a) for LHS of the equation]
x^2 + 12x + 27 = 0
(x + 3)(x + 9) = 0
x = -3 or -9