If \(\frac{3x^2 + 3x - 2}{(x - 1)(x + 1)}\) = P + \(\frac{Q}{x - 1} + \frac{R}{x - 1}\)
Find the value of Q and R
The fraction \( \dfrac{3x^2 + 3x - 2}{(x-1)(x+1)} \) is improper (numerator and denominator both degree 2), so it resolves as \[ \frac{3x^2 + 3x - 2}{(x-1)(x+1)} = P + \frac{Q}{x-1} + \frac{R}{x+1}. \]
Find P. The denominator expands to \( x^2 - 1 \). Divide: \[ 3x^2 + 3x - 2 = 3(x^2 - 1) + (3x + 1), \] so \( P = 3 \) and the proper remainder is \( \dfrac{3x + 1}{(x-1)(x+1)} \).
The fraction \( \dfrac{3x^2 + 3x - 2}{(x-1)(x+1)} \) is improper (numerator and denominator both degree 2), so it resolves as \[ \frac{3x^2 + 3x - 2}{(x-1)(x+1)} = P + \frac{Q}{x-1} + \frac{R}{x+1}. \]
Find P. The denominator expands to \( x^2 - 1 \). Divide: \[ 3x^2 + 3x - 2 = 3(x^2 - 1) + (3x + 1), \] so \( P = 3 \) and the proper remainder is \( \dfrac{3x + 1}{(x-1)(x+1)} \).