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.