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Consider the following function:

$$f(x) = (x+5)^2$$

Its graph will be a parabola only crossing the $x$ $axis$ once.

By setting the equation to $0$ and using our good old $quadratic$ $formula$ gives the same roots repeated.

But why is this so?

Clearly it passes through one point once. And if it did not, it would lose it identity as a function, and would become a multifunction.

Is there any logic behind this or is it just to preserve the theorem that says,

"A polynomial of n degree has exactly n roots"

I'll be thankful to any possible logical explanation.

Sanji Vinsmoke
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  • How do you know that 'clearly it passes through one point once'? How about if the equation passes through two points, but the two points are at the same place? – Toby Mak Dec 02 '17 at 11:19
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    The logic is that quadratic can be written in the form of two factors $c(x-a)(x-b)$. Here $a$ and $b$ are roots. Since quadratic can always be written as two factors, we always get two roots (this is not a proof). No nothing about how many times what crosses what. – jonsno Dec 02 '17 at 11:34

1 Answers1

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You are correct. It is misleading to say it has two roots. By any reasonable, honest and unbiased interpretation, "$p$ has $n$ roots" should mean that the cardinality of the set of roots is $n$.

However, mathematicians have the bias of usefulness and laziness. They can comprehend when they mean that "$p$ has $n$ roots" or that

$p$ has $n$ roots, counted with multiplicity

from context. This last phrase is much more adequate, but mathematicians usually don't see its need since, as I said, they can infer it from context (however, this creates confusion for the uninitiated, as you are experiencing).

So, what means "multiplicity"? One way to interpret the idea (and define the term) is that if $a$ is a root for $p$, you can divide $p$ by $(X-a)$. You are left with a result, $p_1$. It can happen that $p_1$ can then also have $a$ as a root. If so, we can yet again divide by $(X-a)$, getting a result $p_2$. The number of times we can repeat this process until we can't anymore is what we interpret as the "number of times that $a$ is a root of $p$", and call it the multiplicity of $a$ as a root of $p$.

It is easy to see that the multiplicity of a root $a$ of a polynomial $p$ over $\mathbb{C}$ is the exponent of $(X-a)$ on the factorization in linear terms. With a bit of thought, then you get the result that you stated under quotes. More carefully phrased, it states:

The sum of the multiplicities of the roots of a complex polynomial is exactly $n$.

Alternatively,

Every complex polynomial has exactly $n$ roots, when they are counted with multiplicity.

And then, lazily:

Every complex polynomial has exactly $n$ roots.

Aloizio Macedo
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