How to Solve for A, B, and C in x^2+4x-2=A(x+2)(x-2)+Bx(x-2)+Cx(x+2)

[fermio]

Homework Statement

$$x^2+4x-2=A(x+2)(x-2)+Bx(x-2)+Cx(x+2)$$
How to find A, B and C?

Homework Equations

Answer is A=1/2, B=-3/4, C=5/4.

The Attempt at a Solution

A+B+C=1
A(2-2)+B(0-2)+C(0+2)=4
A*2*(-2)+B*0*(-2)+C*0*2=-2

A+B+C=1
0A-2B+2C=4
-4A+0B+0C=-2

C=2+B
A=1/2
1/2+B+2+B=1
2B=-3/2
B=-3/4
C=2-3/4=5/4

 

[J.D.]

I would start by "simplifying" the R.H. side and then compare it to the L.H. side. Thus:

$$A(x+2)(x-2)+Bx(x-2)+Cx(x+2) = Ax^2 -4A +Bx^2 - 2Bx +Cx^2+2Cx = \dots = x^2+4x-2$$

 

[HallsofIvy]

You don't tell how you arrived at those "relevant equations". J.D.'s suggestion, multiply out the right side will work, but this is easier:
Since x= 0, -2, or 2 will make one of the factors on the right 0, let x be each of those in turn;
If x= 0, -2= -4A
If x= 2, 10= 8C
If x= -2, -6= -8B