各位知識朋友 , 入黎幫下手呀! (數學F.2 ) (7)

1. If (x + 6)(x - 6) ≡ Ax^2 + Bx + C , find the value of A,B and C
LHS=(x+6)(x-6)=x^2-36
Compare the coeffecient of x^2, x and the constant
A=1
B=0
C=-36

2. If - 8x^2 + Ax - B ≡ - 4 (Cx^2 + 1) , find the value of A,B and C
RHS=-4Cx^2-4
Compare with the coeffecient of x^2, x and the constant
-4C=-8
C=2
A=0
-B=-4
B=4

3. If A (x - 8)(x + 2) ≡ - x^2 + Bx + C , find the value of A,B and C
LHS=A(x-8)(x+2)
LHS=A(x^2-6x-16)
LHS=Ax^2-6Ax-16A
A=-1
-6A=B
B=-6(-1)=6
-16A=C
C=-16(-1)=16

1.
(x + 6)(x - 6) ≡ Ax^2 + Bx + C
(x + 6)(x - 6)
= x^2 - 6x + 6x + 6^2
= x^2 + 36
equating:
x^2 + 36 ≡ Ax^2 + Bx + C
we got that : A = 1 , B = 0 , C = 36
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2.
- 8x^2 + Ax - B ≡ - 4 (Cx^2 + 1)
put x = 0
-B = -4(1)
B = 4
put B = 4 , x = 1
-8(1)^2 + A(1) - 4 = -4[C(1)^2 + 1]
-8 + A - 4 = -4C - 4
A - 12 + 4 = - 4C
A = 8 - 4C
put A= 8 - 4C , B = 4 , x = -1
-8(-1)^2 + (8 - 4C)(-1) - 4 = - 4 [C(-1)^2 + 1]
-8 - 8 + 4C - 4 = - 4C - 4
4C - 4 - 16 + 4 + 4C = 0
8C = 16
C = 2
put C = 2 into A = 8 - 4C
A = 0
i.e A= 0 , B = 4 , C = 2 .
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3.
A (x - 8)(x + 2) ≡ - x^2 + Bx + C
put x = 8
0 = -(8)^2 + 8B + C
0 = -64 + 8B + C
C = 64 - 8B.............................(1)
put x = -2
0 = -(-2)^2 - 2B + C
0 = -4 - 2B + C........................(2)
put (1) into (2)
0 = -4 - 2B + (64 - 8B)
0 = -4 - 2B + 64 - 8B
4 - 64 = -10B
B = 6
put B = 6 into (1)
C = 64 - 8(6)
C = 16
put B = 6 , C = 16 , x = 1
A[(1) - 8][(1) + 2] = -(1)^2 + 6(1) + 16
A(-7)(3) = -1 + 6 + 16
-21A = 21
A = -1
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i hope this answer can help you .