Find constants and such that
\[\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\]
for all x such that x is not equal to -1 and x is not equal to -2. Give your answer as the ordered pair .
If you let x go infinity, then you get 0 = A + B.
If you let x = 0, then you get -7/2 = A/(-2) + B.
The solution to this system is then A = 7/3, B = -7/3, so (A,B) = (7/3,-7/3).
+26367 Find constants and such that
\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{A}{x - 2} + \dfrac{B}{x + 1}\)
\(\begin{array}{|lrcll|} \hline & \dfrac{x + 7}{x^2 - x - 2} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad x^2 - x - 2 =(x - 2)(x + 1) \\\\ & \dfrac{x + 7}{(x - 2)(x + 1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad \times (x - 2)(x + 1) \\\\ & \mathbf{x + 7} &=& \mathbf{A(x + 1) + B(x - 2)} \\ \hline x=2: & 2 + 7 &=& A(2 + 1) + B(2 - 2) \\ & 9 &=& 3A + 0 \\ & 9 &=& 3A \\ & A &=& \dfrac{9}{3} \\ & \mathbf{A} &=& \mathbf{3} \\ \hline x=-1: & -1 + 7 &=& A(-1 + 1) + B(-1 - 2) \\ & 6 &=& 0 -3B \\ & 6 &=& -3B \\ & B &=& -\dfrac{6}{3} \\ & \mathbf{B} &=& \mathbf{-2} \\ \hline \end{array} \)
\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{3}{x - 2} - \dfrac{2}{x + 1}\)
