maths恆等式題!?

(2x + 3)(4x - 5) ≡ A(3x - 1)(x + 2) + B(x + 1)(x - 1) + C
As it is an identity, so it is satisfied for any values of x.
Put x = 1, we get,
(2 + 3)(4 - 5) = A(3 - 1)(1 + 2) + B(1 + 1)(1 - 1) + C
==> -5 = 6A + C ............... (i)
Put x = 0, we get,
(0 + 3)(0 - 5) = A(0 - 1)(0 + 2) + B(0 + 1)(0 - 1) + C
==> -15 = -2A - B + C
==> 15 = 2A + B - C ......... (ii)
Put x = -1, we get,
(-2 + 3)(-4 - 5) = A(-3 - 1)(-1 + 2) + B(-1 + 1)(-1 - 1) + C
==> -9 = -4A + C
==> 9 = 4A - C .................. (iii)
(i) + (iii), we get,
4 = 10A
==> A = 0.4
Sub. into (i), we get,
-5 = 6(0.4) + C
==> C = -7.4
Sub. into (ii), we get,
15 = 2(0.4) + B + 7.4
==> B = 6.8
∴ A = 0.4, B = 6.8, C = -7.4

Sol
(2x+3)(4x－5)=A(3x－1)(x+2)+B(x+1)(x－1)+C
(8x^2+2x－15)=A(3x^2+5x－2)+B(x^2－1)+C
(8x^2+2x－15)=(3A+B)x^2+5Ax+(－2A－B+C)
3A+B=8，5A=2，－2A－B+C=－15
A=0.4
3*0.4+B=8
B=6.8
－2*0.4-6.8+B=－15
C=－15+7.6=－7.4

（２ｘ＋３）（４ｘ－５）≡Ａ（３ｘ－１）（ｘ＋２）＋Ｂ（ｘ＋１）（ｘ－１）＋Ｃ
left=（２ｘ＋３）（４ｘ－５）
left=8x^2 -10x+12-15
left=8x^2+2x-15
right=Ａ（３ｘ－１）（ｘ＋２）＋Ｂ（ｘ＋１）（ｘ－１）＋Ｃ
right=A(3x^2 +6x-x-2)+B(x^2-1)+C
right=A(3x^2+5x-2)+Bx^2-B+C
right=3Ax^2+5Ax-2A+Bx^x-B+C
right=(3A+B)x^2 +5Ax+(-2A-B+C)
3A+B=8 5A=2 -2A-B+C=-15
3(0.4)+B=8 A=0.4 -2(0.4)-B+C=-15
B=6.8 A=0.4 -0.8-6.8+C=-15
A=0.4 B=6.8 C=-7.6