What is a "double zero" when I am trying to factor a quadratic equation? A writing quadratic equations problem has a double zero of -1.  I need to use this to find my two zeros and for a standard form equation.
 A: There is a very nice relationship between the zeros of a polynomial and the factors of a polynomial.  
If we have a polynomial, and you know, for example, that $2$ is a zero, then $(x - 2)$ is a factor of that polynomial!  That means if you were to factor the polynomial completely, you would see $(x-2)$ somewhere in there.  Similarly, if $-1$ were a zero, then $(x - (-1))$ would be a factor.  If $c$ is a zero, the factor is always $(x - c)$.
That's really great news, because that means if you have a really big polynomial which seems way too hard to factor, you can use the polynomial long division to break it down.
Now, if we know that there is a double zero, like in your case $-1$, that means not only is $(x - (-1))$ a factor [which, if you simplify, equals $(x + 1)$].  But actually, since $-1$ is a double zero, then $(x - (-1))^{2}$ is a factor of the polynomial.
So $(x - (-1))^{2}$, which equals $(x + 1)^{2}$, is a factor.
So multiply this out and get $x^{2} + 2x + 1$.  Now use polynomial long division to divide this into your original polynomial to get the other factors.
A: Look at Wikipedia:

Every polynomial $P$ in $x$ corresponds to a function, $f(x)=P$ (where the occurrences of $x$ in $P$ are interpreted as the argument of $f$), called the polynomial function of $P$; the equation in $x$ setting $f(x)=0$ is the polynomial equation corresponding to $P$. The solutions of this equation are called the roots of the polynomial; they are the zeroes of the function $f$ (corresponding to the points where the graph of $f$ meets the $x$-axis). A number $a$ is a root of $P$ if and only if the polynomial $x−a$ (of degree one in $x$) divides $P$. It may happen that $x−a$ divides $P$ more than once: if $(x−a)^2$ divides $P$ then $a$ is called a multiple root of $P$, and otherwise $a$ is called a simple root of $P$.

