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".
Yes, 1 is correct. Multiplicity is taken from the exponent of the corresponding linear factor. For example, function $g(x) = (x - 4)^{7}$ has a zero at $x = 4$. Its corresponding linear factor is $(x - 4)$. Thus, since the exponent of this linear factor is $7$, we know that the multiplicity of the zero is $7$.
Zeroes with odd multiplicity cross through the $x$-axis, while zeroes with even multiplicity just touch the $x$-axis. Think of an easy example: linear functions have a zero of odd multiplicity and they cross, while quadratic functions can have a zero of even multiplicity if its vertex is just on the $x$-axis so that its barely touching. When you take calculus, you'll learn about this stuff in more detail: it has to do with tangent lines and derivatives!
Best Answer
Each zero has multiplicity 1 in fact. Looking at your factored polynomial: $$-2x^3-x^2+1=(-x)(x+1)(2x-1)$$ The multiplicity of each zero is the exponent of the corresponding linear factor. If we re-write the factorization in the suggestive form: $$-2x^3-x^2+1=-(x)^1(x+1)^1(2x-1)^1$$ The multiplicity of the root -1 is the exponent of the factor $(x+1)$; so it has multiplicity 1. The same applies for the other two roots.