# Why dividing by zero still works

Today, I was at a class. There was a question:

If $x = 2 +i$, find the value of $x^3 - 3x^2 + 2x - 1$.

$x = 2 + i \;\Rightarrow \; x - 2 = i \; \Rightarrow \; (x - 2)^2 = -1 \; \Rightarrow \; x^2 - 4x + 4 = -1 \; \Rightarrow \; x^2 - 4x + 5 = 0$. Now he divided $x^3 - 3x^2 + 2x - 1$ by $x^2 - 4x + 5$.

The quotient was $x + 1$ and the remainder $x - 6$. Now since $\rm dividend = quotient\cdot divisor + remainder$, he concluded that $x^3 - 3x^2 + 2x - 1 = x-6$ since the divisor is $0$.

Plugging $2 + i$ into $x - 6$, we get $-4 + i$.

But my question is, how was he able to divide by zero in the first place? Why did dividing by zero work in this case?

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BTW: in practice, this is one of the most efficient methods to evaluate a polynomial at complex values, if your computing environment does not support complex arithmetic. This is in the same spirit as using synthetic division for polynomial evaluation (Horner's method). –  Ｊ. Ｍ. Jun 1 '13 at 15:21
@J.M. Surprisingly, that is exactly what he also said. –  Parth Kohli Jun 1 '13 at 15:22
Was the topic under discussion the Remainder & Factor Theorems in the complex numbers or was it an exercise in complex arithmetic? I would have had no inkling whatsoever to do anything other than substitute thereby evaluating by substitution. Synthetic division would have been called for if this were an exercise in applying the Remainder Theorem but synthetic division with complex numbers can be messy. If the topic at hand was functions in C, then I would have evaluated f(2 + i). Too often we offer up solutions, some of them elaborate, without knowing or asking what the context is. I would hav –  David Dyer Jun 1 '13 at 15:52
@DavidDyer Remainder and factor theorem: I am just in 9th grade. –  Parth Kohli Jun 1 '13 at 17:31
The remainder of the comment by David Dyer should read" asked first and offered a solution later. –  Alex Becker Jun 1 '13 at 17:32

The division your teacher did is polynomial division. He did not divide by zero; he divided by $x^2 - 4x + 5$.

The long division he did is just an algorithm that allows you to get the following identity:

$$x^3 - 3x^2 + 2x - 1 = (x^2 - 4x + 5)(x+1) + x-6$$

This is an identity involving multiplication, not division. Now, when you plug in $x = 2 + i$, you're not dividing by zero; you're multiplying by zero, which you should agree is allowed.

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Oh, that makes sense. Thanks! –  Parth Kohli Jun 1 '13 at 15:16
(+1) Thanks for the post. I am thinking about that possible Theorem. –  Babak S. Aug 17 at 8:26

The division was a division in the polynomial ring $\mathbb{C}[x]$ by a non-zero polynomial

For example: by dividing $$x^{2}+2x+1$$ by $$x+1$$ which is a non-zero polynomial we get $$(x+1)(x+1)=x^{2}+2x+1$$ since the quotient is $x+1$ and the reminder is $0$. The result is an equality of polynomials, and it is valid if you set any element in the field in those polynomials, even $x=-1$.

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I think your answer is inappropriate for the level of knowledge OP has. It's entirely correct, but referencing polynomial rings over fields is of no help as he likely doesn't even know what a field or a ring are. –  Cameron Williams Jun 1 '13 at 15:24
it's also not essential or even useful for the post. and there is no need to involve $\mathbb{C}$. –  Jonathan Jun 1 '13 at 17:03

because he's only dividing by zero when $x$ takes that value (or its conjugate) otherwise he's really just figuring out what the next thing he'll need to multiply is. Yes, it may look a little dodgy, think of this as a sort of back-of-the-envelope calculation. It's only a means to find out the next step and since in the final expression we no longer have this factor in the denominator, everything works out fine.

Think for example of Factorizing $x^2+7x+12$. We can similarly divide by $x+4$ and get a result of $x+3$ to conclude that $x^2+7x+12=(x+4)(x+3)$. While we can only technically divide by $x+4$ if $x\not=-4$, the process still works out fine.

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$\begin{eqnarray}{\bf Hint}\quad&& f(x) = r(x) + q(x)\, \color{#c00}{g(x)}\quad \text{[e.g. divide } f\ \text{ by }\ g\ \text{ with quotient }\, q, \text{ remainder }\, r\,]\\ \stackrel{\large \color{#c00}{g(a)\,=\,0}}\Rightarrow && f(a) = r(a)\quad \text{by evaluation at } x = a,\text{ using }\ \color{#c00}{g(a) = 0}. \end{eqnarray}$

Because we wrote the "division" as $\ f = q\,\color{#c00}g + r,\,$ not as $\smash[b]{\, \dfrac{f}{\color{#c00}g} = q + \dfrac{r}{\color{#c00}g},\,}$ there is no division by $\,0\,$ when we evaluate at a root of $\,\color{#c00}g.$

Your question has $\,g\,$ quadratic, with root $\,a = 2+i,\,$ The simpler linear case is well-known.

The linear case $\,g = x\! -\!a\,$ has $\, r = f(a),\,$ i.e. $\,f(x)\equiv f(r)\pmod{\!x\!-\!a}\,$ [Remainder Theorem]. Said quivalently $\ x\!-\!a\mid f(x)\!-\!f(a),\,\$ the Factor Theorem,.

For further insight, and a nontrivial application, see the Heaviside cover-up method for evaluating partial fraction decompositions. This does explicitly involve fractions, and, as such, the circumvention of division by zero is more explicit.

More explicitly, we can use division to calculate polynomial derivatives purely algebraically

$$\begin{eqnarray} f'(a) &=\,& \dfrac{f(x)-f(a)}{x-a}\Bigg|_{\large\, x\,=\,a}\\ \\ {\rm i.e.}\ \ \ f'(a) &=\,& q(a)\ \ {\rm where}\ \ f(x)-f(a) = q(x)(x-a)\end{eqnarray}$$

There is no division by zero because the prior linear equation defines a unique polynomial $\,q(x),\,$ and polynomials have no sinularities, i.e. they can be evaluated at any point. This leads to purely algebraic proofs of familiar analytic results, e.g. the double-root test for polynomials.

For a more extreme example of circumvention of division by $\,0,\,$ see this discussion of a proof of Sylvester's determinant identity. Even some professors have mistakenly thought that this proof involves division by zero (a strong testament to the gaps in the exposition of the universal properties of polynomials in many algebra courses). But here, as in the prior examples, there is, in fact, no division by zero, because, before evaluation, one performs valid polynomial operations that eliminate (apparent) singularities. Thus to gain the full universal algebraic power of formal polynomials, one has to (temporarily) forget their anayltic view (as functions). This is easier said than done, since the analytic (functional) bias is so strongly ingrained in our intution.

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