resolve (3x + 2)/(x^2 + 3x + 1) into partial fractions
The fraction is,
(3x + 2)/(x^2 + 3x + 1)
Solving the quadratic x^2 + 3x + 1, gives us: x = (-3 + √5)/2, and x = (-3 - √5)/2
So, out fraction now becomes
(3x + 2)/[ {x – (-3 + √5)/2}{x – (-3 - √5)/2}] =
4(3x + 2)/[{2x – (-3 + √5)}{2x – (-3 - √5)}] = A/{2x – (-3 + √5)} + B/{2x – (-3 – √5)}
4(3x + 2)/[{2x + 3 – √5)}{2x + 3 + √5)}] = A/{2x + 3 – √5)} + B/{2x + 3 + √5)
4(3x + 2) = A{2x + 3 + √5)} + B{2x + 3 – √5)}
2x = -3 – √5
4[(-9 – 3√5)/2 + 2] = A(-3 – √5 + 3 + √5) + B(-3 – √5 + 3 – √5)
2[-9 – 3√5 + 4] = A(0) - 2B(√5)
B = (5 + 3√5)/ √5
B = √5 + 3
2x = -3 + √5
4[(-9 + 3√5)/2 + 2] = A(-3 + √5 + 3 + √5) + B(-3 + √5 + 3 – √5)
2[-9 + 3√5 + 4] = A(2√5) + B(0)
A = (-5 + 3√5)/√5
A = -√5 + 3
The partial fraction then is,
(3x + 2)/(x^2 + 3x + 1) = A/{2x + 3 – √5)} + B/{2x + 3 + √5)
(3x + 2)/(x^2 + 3x + 1) = (-√5 + 3)/{2x + 3 – √5)} + (√5 + 3)/{2x + 3 + √5)