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
Question 1 Report
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
Answer Details
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 \( \dfrac{3x + 1}{(x-1)(x+1)} \).
Find Q and R from \( 3x + 1 = Q(x+1) + R(x-1) \).
At \( x = 1 \): \( 3(1) + 1 = Q(2) \Rightarrow 4 = 2Q \Rightarrow Q = 2 \).
At \( x = -1 \): \( 3(-1) + 1 = R(-2) \Rightarrow -2 = -2R \Rightarrow R = 1 \).
Hence \( Q = 2 \) and \( R = 1 \) (with \( P = 3 \)).
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