已知对于所有实数 $x$，恒有 $5x^3 + x^2 + 27x + A \equiv Bx(x-1)^2 + C(x-1)(x+3)$，求常数 $A$、$B$ 和 $C$。 多项式 $f(x)$ 分别除以 $2x-1$ 和 $x+2$ 的余数分别为 $4$ 和 $-7$。求 $f(x)$ 除以 $2x^2+3x-2$ 的余数。

1. Find A by substituting x = 1

Let $x = 1$:
$5(1)^3 + (1)^2 + 27(1) + A = B(1)(0)^2 + C(0)(4)$
$33 + A = 0 \implies A = -33$

2. Find C by substituting x = 0

Let $x = 0$ (using $A = -33$):
$5(0) + 0 + 0 + (-33) = B(0)(...)^2 + C(-1)(3)$
$-33 = -3C \implies C = 11$

3. Find B by comparing coefficients of x^3

Expanding the right side, the $x^3$ term is $Bx^3$.
Comparing coefficients of $x^3$:
$B = 5$

4. Verify with the x coefficient

Expand: $Bx(x-1)^2 + C(x-1)(x+3) = Bx^3 + (-2B+C)x^2 + (B+2C)x - 3C$.
Coefficient of $x$: $B + 2C = 5 + 22 = 27$ ✓, and coefficient of $x^2$: $-2B + C = -10 + 11 = 1$ ✓.
Thus $A = -33$, $B = 5$, $C = 11$.