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What is a 'zero of multiplicity'?

I looked up this definition on

How many times a particular number is a zero for a given polynomial. For example, in the polynomial function $f(x) = (x – 3)^4(x – 5)(x – 8)^2$, the zero 3 has multiplicity 4, 5 has multiplicity 1, and 8 has multiplicity 2. Although this polynomial has only three zeros, we say that it has seven zeros counting multiplicity.

What is a 'zero of multiplicity' and where do the "seven zeros" come from (I know the "7" comes from adding the three exponents, "4", the hidden "1", and "2" in the equation, but why call them zeroes?)?

As in, I don't understand where the zero part of the term comes in...
(bold emphasis mine).

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up vote 10 down vote accepted

A "zero" of a polynomial is a value of $x$ at which the polynomial, when evaluated, is equal to zero.

The compound phrase "of multiplicity $k$" (where $k$ is a positive integer) modifies "zero" (e.g., "zero of multiplicity 3", "multiplicity 3" and not just 'multiplicity' is modifying "zero"); what it means is that the zero is actually a solution "multiple times".

It is a theorem (called the Factor Theorem) that if $a$ is a zero of the polynomial $p(x)$, then you can write the polynomial $p(x)$ as $p(x)=(x-a)q(x)$; that is, a product. Any zero of $q$ is also a zero of $p(x)$.

We say that $a$ is a zero "of multiplicity $k$" of $p(x)$ if you can write $p(x)$ as $p(x)=(x-a)^kq(x)$, but not as $p(x)=(x-a)^{k+1}q(x)$.

For example, take $p(x)=x^2-2x+1$. Then $x=1$ is a zero of $p(x)$; in fact, since $p(x) = (x-1)^2$, $1$ is a zero "of multiplicity $2$".

Similarly, $p(x)=x^4 - 9x^3 + 30x^2 - 44x + 24$ has $x=3$ and $x=2$ as zeros (plug them in, you get zero: $p(3) = 81 - 243 + 270 - 132 + 24 = 0$, $p(2) = 16 - 72 + 120 - 88 + 24 = 0$). In fact, $p(x) = (x-2)^3(x-3)$, so $3$ is a zero "once" and $2$ is a zero "three times", so $2$ is a zero "of multiplicity three" and 3 is a zero "of multiplicity one".

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this helps, I'm getting there :) – studiohack Jan 4 '12 at 4:57

Recall that a number $r$ is a root or zero of a polynomial $P(x)$ if $P(r) = 0$: i.e., when you plug in $r$, you get zero.

Every root $r$ of $P(x)$ occurs with a certain multiplicity which is the number of times we can factor out $(x-r)$ from $P(x)$. In your example

$f(x) = (x-3)^4 (x-5)(x-8)^2$

the polynomial is conveniently written as a product of distinct linear factors raised to certain powers. These powers are then the multiplicity of the roots of the polynomial, so $3$ occurs with multiplicity $4$, $5$ occurs with multiplicity $1$ and $8$ occurs with multiplicity $2$.

Based on several years of teaching freshman calculus, I can say that this concept is one of the things that university-level teachers think is covered in precalculus mathematics but seems not to be, at least not in a way that makes precalculus students remember / understand it by the time they get to university calculus.

Casting about for a decent, elementary explanation of this material on the web, I found this page. (By contrast, wikipedia does a rather poor job...)

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Wow! Sir , you are back again ? , its been a long time to hear from you, I hope you are fine and will be fine, In order to add something to Prof.Pete's answer, this concept of zero with multiplicity has many good applications like even in Birch Swinnerton dyer conjecture , one connects the rank of underlying Mordell Weil group to the multiplicity of zero of L-function at $s=1$ @Pete.L.Clark – Iyengar Jan 4 '12 at 4:50

In your example, 3 is a zero of $f$, because $f(3) = 0$. That's all it means to call something a zero of a polynomial. If $n$ is some number, and $g(x)$ is a polynomial, we say $n$ is a zero of $g$ if $g(n)=0$.

Now, if we have two polynomials with different zeros, we know the polynomials are different, but how can we tell if $$ f(x) = (x-2)(x-3)$$ and $$ g(x) = (x-2)(x-3)^2$$ are different? They both have 2 and 3 as their roots, so that won't be quite enough to distinguish them. That's where the multiplicity of a root comes in. We say 3 is a zero of multiplicity 1 for $f$, whereas 3 is a zero of multiplicity 2 for $g$.

Another way to look at this, is if we look at $f(x)$ without the factor of $x-3$, namely, the polynomial $$ f'(x) = \frac{f(x)}{x-3} = x-2,$$ we can see $f'(3) = 3-2 = 1 \neq 0$. However, if we look at $$ g'(x) = \frac{g(x)}{x-3} = (x-2)(x-3),$$ we have $g'(3) = 0$. So in some sense, the multiplicity can be thought of as "how many times can we remove the zero from the polynomial, until it's no longer a zero?"

Caution: When I write the divisions above, do not try to plug the zeros in to them. I.e., you cannot say $$ f'(3) = \frac{f(3)}{3-3} = \frac{0}{0},$$ which I claimed was 1 earlier, as you cannot divide by 0.

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A polynomial $\rm\:f(x)\:$ has a root ("zero") $\rm\:r\:$ of multiplicity $\rm\:n\:$ if $\rm\ f(x)\ =\ (x-r)^n\ g(x)\ $ where $\rm\:g(r)\ne 0\:.\:$ Recall by the Factor Theorem that $\rm\:f(r) = 0\ \iff\ x-r\ $ divides $\rm\ f(x)\:.\:$ The multiplicity simply counts how many factors of $\rm\ x-r\ $ occur (the "degree" or "order" of the root $\rm\:r\:$).

Your example $\rm\ (x-3)^4\:(x-5)\:(x-8)^2\ $ has $\ 4 + 1 + 2\ =\ 7\ $ roots (zeros) counting multiplicities since the roots $\ 3\:,\:5\:,\:8\ $ have multiplicity $\rm\ 4\:,\:1\:,\:2\ $ respectively. Note that if we view the roots as a multiset $\rm\ \{3,3,3,3,5,8,8\}\ $ then the multiplicity of a root is simplicity its multiplicity in this multiset, i.e. the number of times that it occurs.

Abhyankar, a master of algebraic geometry, remarks in his charming exposition [1] that

much of algebraic geometry ultimately gets reduced to the following Fundamental Principle (plain or supplemented).

Fundamental Principle. $\ $ The number of roots (or irreducible factors) of a polynomial $\rm\:f(x)\:$ in one variable, counted with their multiplicities (resp. degrees and multiplicities), equals the degree of $\rm\:f(x)\:.\:$

It is certainly the algebraical key to the various 'counting properly.'

I highly recommend reading Abhyankar's paper. It explains simply and beautifully much more than how to algebraically count properly. Indeed, it won various prestigious awards for expository excellence (AMS Lester R. Ford, MAA Chauvenet Prize).

[1] Abhyankar. Historical Ramblings in Algebraic Geometry and Related Algebra
Amer. Math. Monthly 83 (1976), 409-448

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"zero" comes from the fact that those values (3,5,8 in the example) are the values where the polynomial is equal to zero.

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"the values where the polynomial is equal to zero"? What do you mean? this is exactly my hangup, can you explain? – studiohack Jan 4 '12 at 4:45
If you substitute x=3, x=5, or x=8 into the polynomial, you get 0. And if you substitute any other number, you don't get 0. So 3,5,8 are the zeroes of the polynomial. – Ted Jan 4 '12 at 5:09
okay, that explains it... now what about the multiplicity part? – studiohack Jan 4 '12 at 5:12
Polynomials have the property that if $r$ is a zero of the polynomial $p(x)$, then $p(x)$ is divisible by $x-r$. The multiplicity of a zero is how many times you can divide out $x-r$ from the polynomial before the remaining polynomial no longer has $r$ as a zero. For example, take the zero 3 of your polynomial. You can see that it's divisible by $(x-3)^4$, so you can divide out $x-3$ four times from the polynomial (after dividing by $(x-3)^4$, the polynomial that remains no longer has 3 has a zero), so the zero 3 has multiplicity 4. – Ted Jan 4 '12 at 5:34
Another point I just realized from the title of your question: A phrase like "zero of multiplicity 4" should be parsed as "zero (of multiplicity 4)", not as "(zero of multiplicity) 4". In other words, there's no such thing as a "zero of multiplicity". Instead, a polynomial has a zero, and that zero has a particular number associated with it, which is called its "multiplicity". – Ted Jan 4 '12 at 5:42

With polinomials, you can always write it like:
where $x_{1}$, $x_{2}$, $x_{3}$ etc, are the solutions of the polinomial equation (aka roots aka zeros).

E.g. you can always write $x^2-5x+6$ as $(x-2)(x-3)$ because $2$ and $3$ are the solution of the equation $x^2-5x+6=0$.

Now if the polinomial is $x^2-6x+9$, we have that $x_{1}=x_{2}=3$, so we write that
$x^2-6x+9 = (x-3)(x-3) = (x-3)^2$
and we have that $3$ is a "double" solution to the equation, aka, 3 is a zero with a multiplicity of two.

Your original equation can we written as $f(x) = (x-3)(x-3)(x-3)(x-3)(x-5)(x-8)(x-8)$
which has seven terms, and the whole thing has a value of $0$ if and only if some of them are zero. So "zero 3 has multiplicity 4" is another way of saying that "if we set $x$ to $3$, four terms will be $0$"

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