數學問題...請大家幫幫手

2x^2 - 2ax + 3x - 3a = 2x^2 + bx +b

so

-2a+3 = b
-3a = b

then
a = -3
b = 9

(2x + 3)(x - a) = 2x^2 + b(x +1)
2x^2 - 2ax +3x -3a = 2x^2 + bx + b
considering the coefficient of x,
-2a +3 = b---------(1)
considering the constant,
-3a = b---------(2)
substitute (2) into (1),
-2a +3 = -3a
a = -3
substitute a = -3 into (2),
b = 9

(2x + 3)(x - a) = 2x^2 + b(x +1)

2x^2 + 3x - 2ax - 3a = 2x^2 + b(x +1)

兩邊有2x^2, 所以無左
3x - 2ax - 3a = bx +b

抽3x - 2ax的x, 變成(3 - 2a)x
(3 - 2a)x - 3a = bx + b

So, 3-2a = -3a
a = -3

b = 3a
= -3(-3)
= 9