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I'm studying partial fraction decomposition of rational expression. In this video the guy decompose this rational expression:

$$ \frac{3x-8}{x^2-4x-5}$$

this becomes:

$$\frac{3x-8}{(x-5)(x+1)} = \frac{A}{x-5} + \frac{B}{x+1} $$

$$[(x-5)(x+1)]\times \frac{3x-8}{(x-5)(x+1)} = \frac{A}{x-5} + \frac{B}{x+1}\times[(x-5)(x+1)]$$

$$3x-8 = A(x+1) + B(x-5)$$

then he substitutes $x$ with $-1$ this is when I feel something is wrong.

because $x=-1$ is not in the domain of the original function and all the statement below the original function hold true only if they comply within the domain of the original function isn't it? so at the time this guy substitute $x=-1$ the statement become false?

I'm newbie in math, please explain to me with easy understanding and step by step.

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  • $\begingroup$ The last equation you got for A and B indeed comes with the condition that x is not equal to -1 or 5. However, you can also see that if you find A and B that makes the statement valid for all the allowed x, then it will actually also be valid for x = -1 and x = 5. $\endgroup$ – Count Iblis Jul 27 '16 at 6:50
  • $\begingroup$ Possible duplicate of Is this a valid partial fraction decomposition? $\endgroup$ – Hans Lundmark Jul 27 '16 at 9:53
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When you get to the stage $3x-8 = A(x+1) + B(x-5)$, you've forgotten about the partial fractions and all you're trying to do is determine a polynomial identity. i.e: what $A$ and $B$ will make $A(x+1) + B(x-5)$ identical to $3x-8$ for all values of $x$, this includes $x=-1$ and $5$.

Once you've got those values, then you divide both sides of the polynomial identity by $(x+1)(x-5)$ with the added restriction that $x\neq -1, 5$. So this gives you $$\frac{3x-8}{(x+1)(x-5)} = \frac{A(x-5) + B(x+1)}{(x+1)(x-5)} = \frac{A}{x-5} + \frac{B}{x+1}$$

which holds for all $x \neq -1, 5$. (which is what let us do the cancelling bit on the RHS)


Some explanation: the motivation for using $x=-1$ and $5$ is that is efficient in the sense that it makes one of the terms in the identity $0$, speeding up the process of finding the values of the coefficients. If you wanted to, you could use any other two values such as $x=3$ and $4$ to get two simultaneous equations in $A$ and $B$, allowing you to determine their values, just not as efficiently.

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  • $\begingroup$ You have typos. First typo: The denominator on the RHS of your big equation. And secondly the equation is $3x-8=A(x+1)+B(x-5)$. And Then you have wrong values for x. It is $x=-1 $. Very much failures and many upvotes. $\endgroup$ – callculus Jul 27 '16 at 7:01
  • $\begingroup$ @callculus Fixed, thank you. :-) $\endgroup$ – Zain Patel Jul 27 '16 at 7:03
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    $\begingroup$ I will upvote. But I´m really confused that many users who upvote/downvote don´t read even one second the question or the answer. $\endgroup$ – callculus Jul 27 '16 at 7:05
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The last equation you got for A and B indeed comes with the condition that x is not equal to -1 or 5. However, you can also see that if you find A and B that makes the statement valid for all the allowed x, then it will actually also be valid for x = -1 and x = 5, as you can take the limit to these values of both sides of the equation. This amounts to extending the domain of the polynomials by imposing continuity.

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Observe that the equation

\begin{eqnarray} \frac{3x-8}{(x+1)(x-5)}=\frac{A}{x-5}+\frac{B}{x+1} \end{eqnarray}

is valid only for $x\in \mathbb{R}\backslash \{-1,5\}$, as you have rightly pointed out. However, the equation

\begin{eqnarray} 3x-8=A(x+1)+B(x-5) \end{eqnarray}

is valid for all $x\in \mathbb{R}$.

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    $\begingroup$ This is not an answer to the question. It doesn´t explain why in the video the tutor set $x$ equal to $-1$ $\endgroup$ – callculus Jul 27 '16 at 6:53
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    $\begingroup$ @callculus: The question doesn't ask why the tutor set $x$ equal to $-1$. The question asks why it's valid to do so. $\endgroup$ – user14972 Jul 27 '16 at 13:08
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    $\begingroup$ @Hurkyl So what ? $\endgroup$ – callculus Jul 27 '16 at 13:32
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    $\begingroup$ @callculus: I'm sorry, my intent is not clear. This is an answer to the question. I foolishly believed the reason you gave was the reason you thought it was not. $\endgroup$ – user14972 Jul 27 '16 at 13:39
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    $\begingroup$ @Hurkyl Then upvote the answer if you like it. $\endgroup$ – callculus Jul 27 '16 at 13:43

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