What is 'zero of multiplicity' and 'counting multiplicity'?

Question

What is a 'zero of multiplicity' and and 'counting multiplicity'?

I looked up this definition on Mathwords.com:

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).}

Answer 1

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".

Answer 2

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...)

Answer 3

A polynomial $\rm\,f(x)\,$ has a root ("zero") $\rm\,r\,$ of multiplicity $\rm\,\color{#c00}n\,$ if $\rm\ f(x) = (x-r)^{\large\color{#c00}n} g(x),\,$ for $\rm\,g(r)\ne 0.\,$ Recall by the Factor Theorem $\rm\,f(r) = 0 \iff x-r\ $ divides $\rm\, f(x).\,$ The multiplicity $\rm \,\color{#c00}n\,$ of root $\rm\,r\,$ simply counts how many factors of $\rm\ x-r\ $ occur (the "degree" or "order" of the root $\rm\,r$).

Your case $\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

Answer 4

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.

Answer 5

With polinomials, you can always write it like:
$(x-x_{1})(x-x_{2})(x-x_{3})...$
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 $0$"

Answer 6

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

Answer 7

As everyone else has stated, the zero's of a function are simply the x-values where the function evaluates to zero.

The multiplicity of the zero is the degree of the term within the function that evaluates to 0. For example $f(x) = (x-2)^2$ has a multiplicity of $2$ for its zero at $x=2$

Rules of multiplicity: If the multiplicity is even (such as 2) then the graph will only touch the x-axis and then turn around. If the multiplicity is odd, then the graph crosses the x-axis at the zero. Another general rule is that if the multiplicity is greater than 10, the graph tends to flatten out at that zero.