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This may be an incredibly stupid question, but why does partial fraction decomposition avoid division by zero? Let me give an example:


Multiplying both sides by $x(x+1)$ we have:


when $x \neq -1$ and $x \neq 0$.

What is traditionally done here is $x$ is set to $-1$ and $0$ to reveal: $$-3+2=-B \implies 1=B$$ and $$2=A$$

so we find that


Why can $x$ be set equal to the roots of the denominator (in this case, $0$ and $-1$) without creating a division by zero problem?

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If two polynomials in $x$ are equal for infinitely many $x$, then they are equal for all $x$. – GEdgar Jul 1 '12 at 1:17
I would say "traditionally" we equate (coefficients of) like powers of x (in the numerator). After all we need something that works for higher degrees in general. – hardmath Jul 1 '12 at 1:18
up vote 6 down vote accepted

Good question! This is my crude interpretation (see Bill's answer for a shot of rigor)

What is actually being equated is the numerator, not the denominator. So in your example, you have that

$$\frac{{3x + 2}}{{x\left( {x + 1} \right)}} = \frac{A}{x} + \frac{B}{{x + 1}}$$


$$\frac{{3x + 2}}{{x\left( {x + 1} \right)}} = \frac{{A\left( {x + 1} \right) + Bx}}{{x\left( {x + 1} \right)}}$$

if $${3x + 2 = A\left( {x + 1} \right) + Bx}$$

$$3x + 2 = \left( {A + B} \right)x + A$$

which implies

$${A + B}=3$$


which in turn gives what you have.

When we equate numerators we "forget" about the denominators. We're focused in the polynomial equality

$$3x + 2 = \left( {A + B} \right)x + A$$

only. Thought it might be unsettling to be replacing by the roots of the denominators, we're not operating on that, so we're safe.

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Hint $\ $ If $\rm\:f(x),\,g(x)\,$ and $\rm\:h(x)\!\ne\! 0\:$ are polynomial functions over $\rm\:\mathbb R\:$ (or any infinite field) then

$$\rm\begin{eqnarray}\dfrac{f(x)}{h(x)} = \dfrac{g(x)}{h(x)} &\Rightarrow&\rm\ f(x) = g(x)\ \ for\ all\,\ x\in\mathbb R\, \ such\ that\ h(x)\ne 0\\ &\Rightarrow&\rm\ f(x) = g(x)\ \ for\ all\ \,x\in \mathbb R \end{eqnarray}$$

since $\rm\:p(x) = f(x)\!-\!g(x) = 0\:$ has infinitely many roots, viz. all $\rm\:x\in \mathbb R\:$ except the finitely many roots of $\rm\:h(x),\,$ so $\rm\:p\:$ is the zero polynomial, since a nonzero polynomial over a field has only finitely many roots (no more than its degree). Hence $\rm\: 0 = p = f -g\:\Rightarrow\: f = g.$

Thus to solve for the variables that occur in $\rm\:g\:$ it is valid to evaluate $\rm\:f(x) = g(x)\:$ at any $\rm\:x\in \mathbb R,\:$ since it holds true for all $\rm\:x\in \mathbb R\:$ (which includes all real roots of $\rm h).$

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Paying careful attention to the logic of the first step, we are saying that (for a given $A$ and $B$), the equation

$$ \frac{3x+2}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1} $$

holds for all $x \neq 0,-1$ if and only if the equation

$$ 3x+2=A(x+1)+Bx $$

holds for all $x \neq 0,-1$.

Now, if we can find an $A$ and a $B$ so that $3y+2=A(y+1)+By$ holds for all values of $y$, then clearly $3x+2=A(x+1)+Bx $ holds for all $x \neq 0,-1$. So if substituting $y=0$ and $y=-1$ allows us to find $A$ and $B$, then we get a good answer.

Incidentally, a stronger statement is true: the equation

$$ 3x+2=A(x+1)+Bx $$

holds for all $x \neq 0,-1$ if and only if the equation

$$ 3y+2=A(y+1)+By $$

holds for all $y$. So this guarantees that we don't lose any solutions to the former problem when we solve it by instead considering the latter problem.

Aside: if one pays attention to what they mean, one doesn't really need to to introduce a new dummy variable $y$. However, I hoped it might add a bit more clarity if the variable $x$ is always restricted to be $\neq 0,-1$.

It may be useful to note that you use a similar sort of reasoning for limits. e.g. to find the value of

$$ \lim_{x \to 0} \frac{x^2}{x} $$

you observe that $x^2/x = x$ for all $x \neq 0$ so that

$$ \lim_{x \to 0} \frac{x^2}{x} = \lim_{x \to 0} x$$

and then you apply the fact that $x$ is continuous at $0$ to obtain

$$\lim_{x \to 0} x = 0$$

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It's a legit question. The key point is the following lemma

Let $f(x)=p_1(x)/q_1(x)$, $g(x)=p_2(x)/q_2(x)$ be two rational functions. If $f(x)=g(x)$ for all $x$ s.t. $q_1(x)\neq 0$ and $q_2(x)\neq 0$, then $p_1(x)=p_2(x)$ everywhere. This implies that you can get rid of the denominators (for instance multiplying both sides by the least common denominator) and enforcing the equality of functions only in the numerators.

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