Given that 4x^2-6x+9=a(x-1)(2x+1)+b(x-1)+c for all values of x, find the values of a, b and c

4x² - 6x + 9 =  A(x-1)(2x+1)+B(x-1)+C

To eliminate A and B, we need to set (x - 1) = 0   x = 0 + 1   x = 1 

we substitute x = 1 to both sides of the equation

4x² - 6x + 9 =  A(x-1)(2x+1)+B(x-1)+C                    x = 1

4*(1²) - 6*1 + 9 =  A(1-1)(2*1+1)+B(1-1)+C

4 - 6 + 9 = A*(0)(2 + 1) + B*0 + C

7 = 0 + 0 + C

7 = C

C = 7

To eliminate only A, we have to set (2x + 1) = 0

2x + 1 = 0,  2x = 0 - 1,  2x = -1,  x = -1/2 

we substitute x = -1/2, that is x = -0.5 to both sides of the equation

4x² - 6x + 9 =  A(x-1)(2x+1)+B(x-1)+C                    x = 1

4*(-0.5²) - 6*-0.5 + 9 =  A(1-1)(2*-0.5+1)+B(-0.5-1)+C
 
4*(-0.25) + 3 + 9 = A*(0)( ) + B(-1.5) + C

-1 + 3 + 9 = -1.5B + C

11 = -1.5B + C,            But recall C = 7

11 = -1.5B + 7

1.5B = 7 - 11

1.5B = -4

B = -4/1.5 =  -4/(3/2)

B = - 8/3 

Comparing the coefficient of x² on both sides of the equation:

The Ax on the right would multiply with 2x = Ax*2x = 2Ax², this is the only term of x² on the right

4x² on the left        =    2Ax² on the right

4x² = 2Ax²

4 = 2A

2A = 4

A = 4/2

A = 2

Therefore  A = 2,  B = -8/3, and C = 7

Hope this explains it.

Gotta be d the music of previous eras