Is there a reason for some polynomial quotients to have a remainder equals to zero? I was helping some highschool students with factorization exercises. They had alternatives to choose the correct factor. Then one of them said to me:
We use a calculator and evaluate some prime numbers in the quotient by the polynomial and the possible answer, the one with a integer result is the correct.
This seems right, but there's something I don't understand, because, in case of having two options with integer result, they choose the one with lower absolute value. 
Can someone explain me or mention a theorem that validates their argument?
For example:
One factor of $\ 6y-3x^2-6x+3y^2$ is:
a) $x+y$
b) $y-x$
c) $x-y-2$
d) $x+y-2$
They use $x=11$ and $y=7$, then the polynomial $\ 6y-3x^2-6x+3y^2=-240$ and the alternatives are: 
a) $x+y=18$, dividing $-240\div18=-\frac{40}{3}$
b) $y-x=-4$, dividing $-240\div-4=60$
c) $x-y-2=2$, dividing $-240\div2=-120$
d) $x+y-2=16$, dividing $-240\div16=-15$
So finally, they choose as the correct answer the option D).
I know that $\ 6y-3x^2-6x+3y^2=3(y-x)(x+y-2)$, but that's my questions, why using a value on the polynomial they get an integer as the value of the quotient?
 A: Lets suppose $P(x,y) = Q(x,y)R(x,y)$, where $P(x,y)$, $Q(x,y)$, $R(x,y)$ are polynomials with integer coefficients. Then, $P(m,n)$, $Q(m,n)$, $R(m,n)$ are all integers for any choice of integers $m,n$. 
Therefore, "If $Q(x,y)$ is a factor of $P(x,y)$, then $Q(m,n) \mid P(m,n)$ for all integers $m,n$."
The contrapositive to this statement is "If $Q(m,n) \not\mid P(m,n)$ for some integers $m,n$, then $Q(x,y)$ is not a factor of $P(x,y)$." So, if they can find two integers $m,n$ such that $Q(m,n)$ does not divide $P(m,n)$, then they can eliminate the possibility of $Q(x,y)$ being a factor of $P(x,y)$. 
Of course, showing that $Q(m,m) \mid P(m,n)$ for one choice of $m,n$ does not prove that $Q(x,y)$ is a factor of $P(x,y)$. 
If they are allowed to assume that one of the answer choices is correct, and they successfully rule out all other answer choices, then they will get the correct answer. 
However, if they only try one pair of integers $m,n$, and there are still multiple answer choices left, choosing the one that yields the lowest absolute value will not always work. 
Try the following question: "One factor of $4x^2-y^2$ is A) $2x-y$, B) $x+10y$"
If your student tried $x = 7$ and $y = 5$, then he would get $4x^2-y^2 = 171$, $2x-y = 9$, and $x+10y = 57$. Both $9$ and $57$ divide $171$, but since $\dfrac{171}{9} = 19 > 3 = \dfrac{171}{57}$, your student would incorrectly pick $x+10y$ as the factor of $4x^2-y^2$. 
A: Hint $\ y-x\mid 6(y-x)+3(y^2-x^2)\ $ so a correct answer is b).
Re: your method: suppose $\,g\mid f,\,$ i.e. $\,f(x,y) = g(x,y) h(x,y)\,$ for polynomials with integer coeff's.
Then $\ f(m,n) = g(m,n) h(m,n)\ $ so $\,f(m,n)/g(m,n) = h(m,n)\,$ is an integer. Thus, if this fraction is not an integer then $\,g\,$ does not divide $\,f.\,$ For example, if $\,f\,$ is the given polynomial, then evaluating $\,f/g\,$ at $\,x=2,\ y = 5\,$ for each of the listed possible factors $\,g\,$ yields the values $\, 81/7,\ 27, -81/5,\ 81/5.\ $ This excludes all but the second choice b) as a possible factor.
Remark $\ $ This divisibility test works because polynomial evaluation is a ring homomorphism, so preserves divisibility. More generally suppose $\,h: R \to R'\,$ is a $\rm\color{#c00}{multiplicative}$ map between rings (or multiplicative monoids), i.e. $\,\color{#c00}{h(rs) = h(r)h(s)}\,$ for all $\,r,s\in R.\,$ Then $\,b\mid a\,\Rightarrow\, h(b)\mid h(a)\ $ since
$$b\mid a\,\Rightarrow\ bc = a\,\Rightarrow\,  \color{#c00}{\underbrace{h(bc)}_{\large h(b)\,h(c)}}\!\! = h(a)\,\Rightarrow\, h(b)\mid h(a)\qquad$$
If divisibility testing in $R'$ is simpler than that in $R,\,$ then this yields a useful divisibility criterion.
For example, consider invertibles (units). If $\,ab = 1\,$ then $\,h(a)h(b) = 1,\,$
 i.e. $\,a\mid 1\,\Rightarrow\, h(a)\mid 1.\,$ Therefore:  $\,a\,$ invertible $\,\Rightarrow\,h(a)\,$ invertible. When $\,a = A\,$ is a square matrix over a field and $\,h\,$ is the (multiplicative) determinant map, this specializes to: $\,A\,$ invertible $\,\Rightarrow\, \det(A)\ne 0.\,$ Hence this well-known linear algebra result is a special case of a general divisibility criterion.
