F.2 Maths Problem

LHS
= A(X + 3)^2 + B(X + 3)(X - 4)
= A [X^2 + 6X + 9] + B[X^2 - X - 12]
= (A + B) X^2 + (6A - B) X + (9A - 12B)

Comparing cofficients with RHS, we have
A + B = 1 ... [1]
6A - B = 13 ... [2]
9A - 12B = C ... [3]
[1]+[2], 7A = 14, A = 2
Put A = 2 into [1], 2 + B = 1, B = -1
Put A = 2 and B = -1 into [3],
C = 9(2) - 12(-1) = 18 + 12 = 30
So, A = 2, B = -1, C = 30

Check:
LHS
= 2(X + 3)^2 - (X + 3)(X - 4)
= 2(X^2 + 6X + 9) - (X^2 - X - 12)
= 2X^2 + 12X + 18 - X^2 + X + 12
= X^2 + 13X + 30
RHS = X^2 + 13X + 30 = LHS

A(X+3)^2 + B(X+3)(X-4) = X^2 + 13X + C
put X = -3, 0 = 9 - 39 + C, C = 30
put X = 4, 49A = 16 + 52 + 30, A = 2
2(X+3)^2 + B(X+3)(X-4) = X^2 + 13X + 30
put X = 0, 18 - 12B = 30, B = -1


2007-10-23 21:35:58 補充：
oh, 慢左, 不過用 代入數值既方法 比較快同方便,好過上面乘開條式配對既方法所以, 請投我一票啊, 唔該

A(X+3)^2 + B(X+3)(X-4) = X^2 + 13X + C
A(X^2+6X+9) + B(X^2+3X-4X-12) = X^2 + 13X+ C
(A+B)X^2 + (6A-B)X+9A-12B = X^2 + 13X+ C
Hence, A+B = 1, 6A-B = 13 and 9A-12B = C
Consider
A+B = 1 ...(1)
6A-B = 13 ...(2)
(1) + (2): 7A = 14
A = 2
B = 1-2 = -1
C = 9(2) - 12(-1) = 18-(-12) = 30