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
Find the value of Q and R
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Correct Answer: Option
Explanation:
\(\frac{3x^2 + 3x - 2}{(x - 1)(x + 1)}\) = P + \(\frac{Q}{x - 1} + \frac{R}{x - 1}\)
Long division
\(\frac{6x^{-2}}{(x - 1)(x + 1)}\)
= \(\frac{Q}{x - 1} + \frac{R}{x + 1}\)
6x - 2 = Q (x + 1) + R(x - 1)
Let x = -1
6(-1) - 2 = Q(-1 + 1) + R (-1 - 1)
-6 - 2 = -2R
\(\frac{-8}{-2} = \frac{-2R}{-2}\)
R = 4
From 6x - 2 = Q(x - 1) + R(x - 1)
Let x = 1
6(1) - 2 = Q(1 + 1) + R(1 - 1)
6 - 2 = 2Q
\(\frac{4}{2} = \frac{2Q}{2}\)
= Q = 2
\(\frac{3x^2 + 3x - 2}{(x - 1)(x + 1)}\) = P + \(\frac{Q}{x - 1} + \frac{R}{x - 1}\)
Long division
| \(x^2 - 1\) | 3 |
| 3x\(^2\) + 3x - 2 -(3x\(^2\) - 3x)\(\overline{6x - 2}\) |
\(\frac{6x^{-2}}{(x - 1)(x + 1)}\)
= \(\frac{Q}{x - 1} + \frac{R}{x + 1}\)
6x - 2 = Q (x + 1) + R(x - 1)
Let x = -1
6(-1) - 2 = Q(-1 + 1) + R (-1 - 1)
-6 - 2 = -2R
\(\frac{-8}{-2} = \frac{-2R}{-2}\)
R = 4
From 6x - 2 = Q(x - 1) + R(x - 1)
Let x = 1
6(1) - 2 = Q(1 + 1) + R(1 - 1)
6 - 2 = 2Q
\(\frac{4}{2} = \frac{2Q}{2}\)
= Q = 2