Why does $x^2 = 0$ give a repeated root of $x=0,0$ instead of a single root of $x=0$. For an example function: $x^4-4x^3+4x^2=0$
We can reduce it to   $ x^2(x^2-4x+4)=0$
The function in the parentheses will give the repeating root of $x=-2,-2$.
This makes sense.
However the answer to this function gives the root(s) of $x^2=0$ as $x=0,0$.
I understand the need to have repeated roots, one positive, one negative, for a non-zero whole number, but why do I need two roots for zero? Do I just pretend to have  $+0,-0$ as my roots?
 A: You can factor $x^2=(x-0)(x-0)$.
A: It may seem superfluous, but there are sound reasons for defining multiplicity of roots the way we do.  A function behaves differently in the neighborhood of a single root than it does in the neighborhood of a double root—the graphs of $f(x) = x$ and $g(x) = x^2$ illustrates that.  The fact that the root happens to be $0$ doesn't affect that.
It makes better sense when making generalizations about limits, derivatives, and the like, if we retain the information about the multiplicity of roots.
A: 
I understand the need to have repeated roots, one positive, one negative, for a non-zero whole number.  

This is wrong. Repeated roots are for the same number, not "one positive, one negative". We can see this in your example of $x^2(x-2)^2=0$, where there is a repeated root of $x=2$. Following from this, we can just as easily tell that $x=0$ is another repeated root (a good way to convince yourself is to write $x^2$ as $(x-0)^2$).
Now what do repeated roots mean? If we graph the function $f(x)=x^4-4x^3+4x^2$ 
then it is clear that the repeated roots of $f(x)$ account for turning points on the line $y=0$.

If you are familiar with calculus, you can prove that at repeated roots the derivative is $0$. Consider the polynomials $P(x)$ and $g(x)$, where
$$P(x)=(x-a)^2g(x)$$
Then, a repeated root is $x=a$, and
$$P'(x)=2(x-a)g(x)+(x-a)^2g'(x)$$
$$\implies P'(a)=0$$
A: You seem to be a bit confused by the meaning of "repeated" when we say "repeated root."
The equation $$x^2 - 4 = 0$$ does not have any repeated roots.  The solution set is $x \in \{-2, 2\}$ and these roots are distinct--each root has multiplicity $1$.
The equation $$x^2 - 4x + 4 = 0$$ has one root, $x = 2$, but it is a repeated root with multiplicity $2$.
The equation $$x^3 + 3x^2 + 3x + 1 = 0$$ has one root, $x = -1$, and it is a repeated root with multiplicity $3$.
To understand what's going on, let's factor each of the above polynomials:  $$x^2 - 4 = (x-2)(x+2) \\ x^2 - 4x + 4 = (x-2)(x-2) \\ x^3 + 3x^2 + 3x + 1 = (x+1)^3.$$  Now we see that:


*

*The number of distinct roots is at most equal to the polynomial degree.

*If all roots are distinct, the number of roots is equal to the polynomial degree.

*The maximum multiplicity of any repeated roots cannot exceed the polynomial degree.

*The sum of the multiplicities of all distinct roots equals the polynomial degree.


The last statement above is essentially the Fundamental Theorem of Algebra.
In general, a polynomial of degree $n$ with coefficients in $\mathbb C$ will admit a factorization into exactly $n$ linear factors:  $$\sum_{k=0}^n a_k z^k = a_n \prod_{j = 1}^n (z - \zeta_j),$$ for $\{a_k\}_{k=0}^n \in \mathbb C$ and $a_n \ne 0$, where $\{\zeta_j\}_{j = 1}^n \in \mathbb C$ is the set of roots with multiplicity; that is to say, the set may contain distinct and repeated elements.  This factorization is always possible in $\mathbb C$.
A: Let $$f (x )=x^4-4x^3+4x^2$$
then
$$f'(x)=4x^3-12x^2+8x $$
$$=4x (x^2-3x+2)=4x (x-2 )(x-1)$$
$0$ is a root of $f (x)=0$ and of $f'(x)=0$ thus it is a double root, with order of multiplicity $2$.
$2$ is also a double root.
