maths問題。趕!!!

3(x²-1)+Ax+15≡(B-x)(Cx-2)
3x²-3+Ax+15 ≡ -Cx²+(BC+2)x-2B
3x²+Ax+12 ≡ -Cx²+(BC+2)x-2B

Compare the coefficient of x², x and constant
C = -3

-2B = 12
B = -6

A = BC + 2
A = 20

3(x²-1)+Ax+15≡(B-x)(Cx-2)

Left hand side:
3 (x²-1) + Ax + 15
= 3x²-3 +Ax + 15
=3x² +Ax +12

Right Hand side:
(B-x)(Cx-2)
= BCx - 2B-Cx²+2x
= - Cx²+(BC+2)x-2B

Compare the co-effficent of the like term:
so -C=3,
=> C=-3
and -2B=12,
=> B=-6
and A=BC+2
=> A=(-6)(-3) +2
=> A= 18+2
=> A= 20

The answer: A= 20, B= -6, C= -3

This is an identity, therefore the coefficents of x² term, x term and constant term are equal at two sides.

For the Left hand side,
3(x²-1) + Ax + 15
=3x² + Ax +12

For the Right hand side,
(B-x) (Cx-2)
= -Cx² + (BC+2)x - 2B

Compare the coefficient of x²
3 = -C
C = -3

Compare the constant term
12 = -2B
B = -6

Compare the coefficient of x
A = BC+2
A = (-3)(-6) + 2
A = 20

Subst. A=20 , B=-6, C=-3 into the identity 3(x²-1)+Ax+15≡(B-x)(Cx-2)
LHS
3(x²-1) + 20x + 15
= 3x² + 20x +12

RHS
(-6-x)(-3x-2)
=3x²+20x+12

First of all, this is a polynomial identity:

3(x²-1)+Ax+15≡(B-x)(Cx-2)
3x²-3+Ax+15≡BCx-2B-Cx²+2x (Expanding right hand side to a quadratic polynomial)
3x²+Ax+12≡-Cx²+(BC+2)x-2B (Rewriting in descending order of x)

Equating the coefficients of each power of x on both sides, we have:
3 = -C ---> C = -3 (Coefficient of x²)
A = BC + 2 ---> (*) (Coefficient of x)
12 = -2B ---> B = -6 (Constant term)

Substituting B = -6 and C = -3 into (*), we have:
A = (-6)(-3) + 2 = 20

L.H.S. = 3(x²-1)+Ax+15
= 3x² - 3 + Ax + 15
= 3x² + Ax + 12
R.H.S. = (B-x)(Cx-2)
= BCx - 2B - Cx² + 2x
= - Cx² + BCx + 2x - 2B
= - Cx² + (BC + 2)x - 2B

∵ L.H.S. ≡ R.H.S.
∴ -Cx² = 3x² ... x² 的系數
-C = 3
C = -3
-2B = 12 ... 常數項
B = 12/-2
B = -6
Ax = (BC + 2)x ... x 的系數
A = (-6)(-3) + 2 ... 代入 B = -6 及 C = -3
A = 18 + 2
A = 20