Partial fractions, disagreement with Wolfram Alpha On my math homework, I have this problem, and WolframAlpha says that $A=-\frac{1}{7}$ and $B=\frac{1}{7}$. However, while solving the problem on my own, I found a restriction that $A \neq -B$. How is it that this answer works? The problem in question is:
$$\dfrac{1}{x^2 - 3 x - 10} = \dfrac{A}{x+2} + \dfrac{B}{x - 5}$$

To find $A$ and $B$:
Note that
$(x + 2)(x - 5) = x^2 -3x - 10; \tag 1$
then from
$\dfrac{1}{x^2 -3x - 10} = \dfrac{A}{x + 2} + \dfrac{B}{x - 5} \tag 2$
we have
$A(x - 5) + B(x + 2) = \dfrac{(x + 2)(x - 5)}{x^2 -3x - 10} = 1; \tag 3$
this may be written
$(A + B)x + (2B - 5A) = 1, \tag 4$
whence
$A + B = 0 \Longrightarrow B = -A, \tag 5$
so that
$-7A = 2(-A) - 5A = 2B -5A = 1 \Longrightarrow A = -\dfrac{1}{7}, \; B = \dfrac{1}{7}; \tag 6$
we check:
$\dfrac{A}{x + 2} + \dfrac{B}{x - 5} = -\dfrac{1}{7}\dfrac{1}{x + 2} + \dfrac{1}{7}\dfrac{1}{x - 5} = \dfrac{1}{7} \left ( \dfrac{1}{x - 5} - \dfrac{1}{x + 2} \right )$
$= \dfrac{1}{7}\left ( \dfrac{x + 2}{x^2 -3x - 10} - \dfrac{x - 5}{x^2 -3x - 10} \right ) = \dfrac{1}{7}\dfrac{7}{x^2 -3x - 10} = \dfrac{1}{x^2 -3x - 10}. \tag 7$
