What is 'zero of multiplicity'?

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".

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...)

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