Integer polynomial and division by binomial If I have an integer coefficient polynomial $f(x)$ and know that it has 2 integer roots, $x_1$ and $x_2$, is it true that
$$f(x) = (x-x_1)(x-x_2) h(x)$$
where $h(x)$ is another integer coefficient polynomial? What is this result called if it is true? How can I prove that?
 A: Let's assume that $x_1,x_2$ are distinct integer roots of integer polynomial $f(x)$.  In that case we can first divide $f(x)$ by $(x-x_1)$, then divide the resulting quotient by $(x-x_2)$ to get the desired result.  So the hard part of the proof is showing that the first quotient is again an integer polynomial:
$$ f(x) = (x-x_1)g(x) $$
Since $x_1\neq x_2$, in order for $f(x_2) = 0$, it must be that $g(x_2) = 0$.  This allows us to do the "division" by $(x-x_2)$ just as we will show the first step, and if $g(x) = (x-x_2) h(x)$, then:
$$ f(x) = (x-x_1)(x-x_2)h(x) $$
With that in mind we will go deeper into the details of why $g(x)$ as the first quotient is an integer polynomial.

The topic of synthetic division is generally covered in Algebra II for high school students where the divisor is binomial $x-r$ and the dividend $f(x)$ is an integer polynomial.  It can be generalized to some extent, but this is already the case our immediate purpose requires.
The computation is equivalent to carrying out a long division of polynomials, but the special form of the divisor $x-r$ allows for less writing and less arithmetic.  Further the equivalence of the synthetic division and the long division allow us to realize that the final remainder $d$ of the computation:
$$ f(x) = (x-r)g(x) + D $$
is exactly $D = f(r)$, the evaluation of polynomial $f(x)$ at $x=r$.
The main idea is to "attack" the leading coefficient of $f(x)$ by subtracting a multiple of $(x-r)$ that eliminates that term, thus reducing the degree of the (partial) dividend.  Suppose that $f(x)$ is an integer polynomial of degree $n$:
$$ f(x) = A_0 x^n + A_1 x^{n-1} + \ldots + A_{n-1}x + A_n $$
Then we could subtract $(x-r)\cdot A_0 x^{n-1}$ from $f(x)$ and get a "new" dividend of degree (at least) one less:
$$ f(x) - (x-r)\cdot A_0 x^{n-1} = (A_1 + rA_0) x^{n-1} + A_2 x^{n-2} + \ldots + A_{n-1}x + A_n $$
We continue to do this subtraction of multiples of $(x-r)$ times the indicated coefficient of smaller and smaller powers of $x$ until at last we have removed all but a final constant remainder $D$.  Piecing everything together we have $f(x) = (x-r)g(x) + D$ as promised, where the degree of the combined quotient $g(x)$ is one less than the degree of $f(x)$.
Notice that although division is being carried, the adjustments of the coefficients at each stage are actually additions of integers.  So by induction not only is the final remainder $D$ an integer, so are the coefficients of the pieced together quotient $g(x)$.
This in brief establishes that if $r$ is a root of $f(x)$, then the final remainder $D$ is zero (because $D=f(r)$ as explained earlier), and we have our factorization of integer polynomials:
$$ f(x) = (x-r)g(x) $$
Let me know if this much detail is not enough, or if more explanation is needed to clarify what has already been sketched.
A: Since $f$ is an integer coefficient polynomial that has $2$ integer roots $x_1$ and $x_2$, it is divisible by $(x-x_1)$ and $(x-x_2)$, meaning that it can be rewritten as $(x-x_1)(x-x_2)h(x)$, where $h(x)$ is another integer coefficient polynomial. This is because $(x-x_1)(x-x_2)$ is a monic polynomial, so it doesn't change the leading coefficient of $f(x)$. This result is a factorization of $f(x)$. 
A: Simply, we have
$$
h(x)=\frac{f(x)}{(x-x_1)(x-x_2)}\in\mathbb{Q}[x]
$$
Therefore, there is a $q\in\mathbb{Z}$ so that $qh(x)\in\mathbb{Z}[x]$ and the content of $qh(x)$ is $1$. By Gauss' Lemma, $(x-x_1)(x-x_2)qh(x)=qf(x)$ has content $1$. Since $f\in\mathbb{Z}[x]$, $q$ divides the content of $qf(x)$, so $q=1$. That is, $h(x)\in\mathbb{Z}[x]$.
A: Consider the polynomial $g_1(x)=f(x+x_1)$. Its coefficients are integer because to compute them you only need to multiply and add coefficients of $f$ and $x_1$.
Since $f(x_1)=0$, it follows that $g_1(0)=0$. This is, the coefficient of $g$ of degree zero is zero. Therefore, $g_1(x)=xg_2(x)$. The coefficients of $g_2$ are the same as the coefficients of $g_1$, just shifted to one degree less. Hence, they are also integers.
Replacing $x$ by $x-x_1$ in the equation $g_1(x)=xg_2(x)$ we get $$f(x)=g_1(x-x_1)=(x-x_1)g_2(x-x_1)$$
The coefficients of $g_3(x)=g_2(x-x_1)$ are integers because, as before, they are obtained from those of $g_2$ by doing only additions and multiplications of them with $x_1$. 
Repeating the same argument with $x_2$ and $g_3(x)$ you get the result.

Aside: Above there were two moments in the argument in which one needs to write polynomials like $f(x-x_1)$ back into powers of $x$, too get its coefficients.
This consists in opening all parentheses. However, this can be done, and faster, by just evaluating $f(x)$ at $x_1$. See Ruffini's rule.
Good moral to remember for other problems: Dividing polynomials by $x-a$ doesn't require division, only evaluation at $a$.
