Consider this partial fraction. $$\dfrac{x-25}{x^2+5x-24}=\dfrac{A}{x-3}+\dfrac{B}{x+8}$$ Multiply both sides by the quadratic $x^2+5x-24$. $$x-25=A(x+8)+B(x-3)$$ From here, I've seen many textbooks set $x=3$, solve for $A$, then set $x=-8$ and solve for $B$. But how come we can do that if we assumed $x\neq3$ and $x\neq-8$ when dividing the quadratic?
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1See e.g. https://math.stackexchange.com/questions/1313454/is-this-a-valid-partial-fraction-decomposition or https://math.stackexchange.com/questions/4001333/why-does-this-partial-fraction-decomposition-work-even-with-division-by-0. – Minus One-Twelfth Nov 15 '23 at 18:25
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@FrankW But in that example, you are assuming that the numerators are equal by multiplying and cancelling the denominators, in which case it is 0/0. – Anirudh Yamunan Govindarajan Nov 15 '23 at 18:29
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3I would frame it as: If $x-25=A(x+8)+B(x-3)$ for all $x$ then $$\dfrac{x-25}{x^2+5x-24}=\dfrac{A}{x-3}+\dfrac{B}{x+8}$$ for all $x \neq 3,8$. So, you can do whatever you need to do to solve the above equation for $A$ and $B$, and if it makes the equation true for all $x$, you can proceed. – Jair Taylor Nov 15 '23 at 18:31
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The reason this works is because of something called the Residue Theorem in complex analysis. – Leonidas Lanier Nov 15 '23 at 18:36
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You can use it to find A and B since you have transformed your equations, and the linear terms in x are no longer in the denominator, so you can treat the newly obtained equation as a separate new equation and thus disregard the assumptions of the previous equation. In multiplying the equation by $ x²+5x-24 $, you have slightly changed the working of the equation; the newly obtained equation works for all real x, including -8 and 3.
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