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.
