proof $z \mid b$ and $w\mid b$ Question I'm working on: Let $a,b$ be integers with $b$ not equal to $0$. suppose $x^2+ax+b=0$ and $x=z,w$. If $z,w$ are integers, show that $z\mid b$ and $w\mid b$.
Is it sufficient for me to show that $x^2+ax+b=(x-z)(x-w)$ so $b=zw$?

Here is a better approach:
We need to show $z|b$ and $w|b$.  What does it mean for $z$ to "divide" $b$?  It means there is some integer $k$ such that $z \cdot k = b$.  Well, what are we given?  We are given that if $x = z$ or $x = w$, then $x$ satisfies $x^{2} + ax + b = 0$.
But that means $z^{2} + az + b = 0$, or $z^{2} + az = -b$.  But then that means $z(z + a) = - b$.  Since $z + a$ is an integer (because $z$ and $a$ are integers), then we have found an integer $k$ such that $z \cdot k = -b$, so $z$ divides $-b$.  Is it clear that if $z$ divides $-b$, then $z$ divides $b$?
You can use the same argument for $w$.  Does this argument make sense to you?  Do you understand every step?
