What do polynomials solve for? Roots? I had a question in which I’ve been hung up over on. I understand if we had a graph x-y plane and we drew points that intercept the $x$-axis at $2$ and $3$, we would write a quadratic equation that satisfies our condition as $(x+3)(x+2)$, or $x^2+5x+6$, and set this equal to $0$ using the null factor law. Then, we would solve it by factoring and would eventually get back to $(x+3)(x+2)=0$.
And we would get $x=-2,-3$ but how come these are the roots(points that intercept the $x$-axis), when we first created this quadratic equation using the points $2$ and $3$ on $x$-axis? I want to understand in short a logical explanation what we are solving for when we are factoring polynomials? 
 A: The roots of the polynomial $x^2-5x+6$ correspond to the values of $x$ where the function $f(x)=(x-3)(x-2)$ is equal to $0$.
They are points of the $x$-axis : $(3,0)$ and $(2,0)$, i.e. the intersection points of the graph of $f(x)$ with the $x$-axis.
A: You are mixing a few different things. 
You seem to want to represent a graph with a polynomial. This is clearly possible, at least on a given finite intervall, if the graph is behaving correctly (for instance, for a given $x$, you have only one $y$). This is indeed one of the main use of polynomials, since they are "simpler" to deal with than many other functions. 
But a graph cannot be "simply" represented by a polynomial that just happen to share the same roots. By the way, a polynomial that has $2,3$ as roots would be $(x-2)(x-3)$ (which explains why you did not fall back on the same roots in your example).
We call roots (by definition) the $x$ such as a function $f(x)$ is equal to $0$ (in short, the solutions to the equation $f(x)=0$).
Roots are not limited to polynomials by the way. You can search for roots for non-polynomials. 
Factoring a polynomial, on the other hand, is a great way to find out what the roots of this polynomial are.
A: The graph of a function $y=f(x)$ intersects the $x$-axis when $y=0$, that is when $f(x)=0$.
Now if a factor $(x-a)$ appears in your polynomial, then the polynomial is zero when $x-a = 0$, which occurs precisely when $x=a$. Therefore
$$\begin{align*}
(x-a)(x-b) = 0 & \Leftrightarrow x-a=0 \text{ or } x-b = 0\\
& \Leftrightarrow x=a \text{ or } x=b
\end{align*}$$
So the graph $y=(x-a)(x-b)$ intersects the $x$-axis when $x=a$ or $x=b$.
This explains the $-$ signs.
So the quadratic $(x+3)(x+2)$ has roots $-3$ and $-2$, whereas the quadratic $(x-3)(x-2)$ has roots $3$ and $2$.
