+743 Find constants A and B such that
(x + 17)/(x^2 - x - 2) = A/(x - 2) + B/(x + 1)
for all x such that $x \neq -1$ and $x \neq 2$. Give your answer as the ordered pair (A,B).
+399 This is pretty simple partial fraction decomposition.
Our first indication that partial fraction decomposition works is \((x-2)(x+1)={x}^{2}-x-2\)
\(\frac{x+17}{{x}^{2}-x-2}=\frac{A}{x-2}+\frac{B}{x+1}\), and we make the denominators equal, \(\frac{x+17}{{x}^{2}-x-2}=\frac{A(x+1)}{(x+1)(x-2)}+\frac{B(x-2)}{(x+1)(x-2)}\), and \(x+17=A(x+1)+B(x-2)\). To solve, we plug in some special values, like 2, so 19 = 3A, A = 19/3. Similarly, by plugging in -1, we get B = -16/3. So out answer is A = 19/3, B = -16/3.\(\)
+399 This is pretty simple partial fraction decomposition.
Our first indication that partial fraction decomposition works is \((x-2)(x+1)={x}^{2}-x-2\)
\(\frac{x+17}{{x}^{2}-x-2}=\frac{A}{x-2}+\frac{B}{x+1}\), and we make the denominators equal, \(\frac{x+17}{{x}^{2}-x-2}=\frac{A(x+1)}{(x+1)(x-2)}+\frac{B(x-2)}{(x+1)(x-2)}\), and \(x+17=A(x+1)+B(x-2)\). To solve, we plug in some special values, like 2, so 19 = 3A, A = 19/3. Similarly, by plugging in -1, we get B = -16/3. So out answer is A = 19/3, B = -16/3.\(\)
