# 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?

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Can you give an example? I'm not sure I follow the exact reasoning. – Matt B. Jul 21 '14 at 22:21
Do you have an example? – johannesvalks Jul 21 '14 at 22:21
I think a little bit more context would help us understand what you are trying to ask. – Mark B Jul 21 '14 at 22:24
Your factorization is incorrect. $6y-3x^2-6x+3y^2=3(y-x)(x+y+2)$ and the correct answer is b. They got an integer there, too, so don't yet know whether it is b or d. – Ross Millikan Jul 21 '14 at 23:39
The method they choose is incorrect. For example, if they are given $15x^{2}+14y^{2}+31xy$ (can be factored as $(30x+42y)(\frac{x}{2}+\frac{y}{3})$) and ask if $30x+42y$ is a factor, they plugging $x=1,y=0$ give $\frac{15}{30}=\frac{1}{2}$, so they end up eliminating the correct answer. Now as you notice, you can "rescale" the 2 factors above by rational ratio so that all the coefficient are integer. Similarly, a modification need to be added to this method: you need to make sure that the ratio cannot be rescaled before you eliminate that answer. Here rescale by $2$ give $15$ and $1$. – Gina Jul 22 '14 at 1:39

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 $4x^2-y^2$.

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I think you can remove the assumption that the factors have integer coefficient. Using Gauss's lemma (over UFD) guaranteed it, in some sense. If the calculated value of $Q$ is $n$ and the calculated value of $P/Q$ is $\frac{a}{b}$ in reduced form then we must have $b|n$ if $Q$ are to be a factor. – Gina Jul 22 '14 at 1:28
@JimmyK4542 : But what if they try with several pairs of primes and with all of them, they get an integer solution? – hkviktor Jul 22 '14 at 18:14

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.

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