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

To find the values of Q and R, we need to first simplify the right-hand side of the equation and then equate the numerators of both sides.

First, we need to find the common denominator of the fractions on the right-hand side:

P + \(\frac{Q}{x - 1}\) + \(\frac{R}{x + 1}\) = \(\frac{Px^2 - P + Q(x + 1) + R(x - 1)}{(x - 1)(x + 1)}\)

Next, we can equate the numerators of both sides:

3x^2 + 3x - 2 = Px^2 - P + Q(x + 1) + R(x - 1)

Now, we need to solve for Q and R by choosing appropriate values of x.

To find Q, we can choose x = 1:

3(1)^2 + 3(1) - 2 = P(1)^2 - P + Q(1 + 1) + R(1 - 1)

4 = P + 2Q

To find R, we can choose x = -1:

3(-1)^2 + 3(-1) - 2 = P(-1)^2 - P + Q(-1 + 1) + R(-1 - 1)

-2 = P - 2R

We now have two equations with two variables (P and Q), and we can solve for Q by substituting the expression we found for P in the previous step:

4 = (3 + 3 - 2P) + 2Q

2Q = 2P + 1

Q = P + \(\frac{1}{2}\)

To find R, we can substitute the expression we found for P and the expression we found for Q in the equation we found for R:

-2 = (3 + 3 - 2P) - 2(P + \(\frac{1}{2}\))

-2 = 6 - 4P

P = \(\frac{4}{3}\)

Substituting this value of P in the expression we found for Q, we get:

Q = \(\frac{4}{3}\) + \(\frac{1}{2}\) = \(\frac{11}{6}\)

Finally, substituting the values of P and Q in the expression we found for R, we get:

-2 = 6 - 4(\(\frac{4}{3}\)) - 2(\(\frac{11}{6}\))

R = -\(\frac{5}{6}\)

Therefore, the value of Q is \(\frac{11}{6}\) and the value of R is -\(\frac{5}{6}\).