Compare the coefficients of $x^3$, $x^2$, $x$, and the constant term from both expressions. What equations do you get for $m$, $nq$, and $a$?

From the constant terms you get $nq=17$. Since 17 is prime and $n$ and $q$ are integers, what are the only possible integer values of $n$?

Use $m=2$ and the fact that $n$ must be a divisor of 17. For each choice, ask: does this expression have to be an integer for all allowed values of $n$ (and all integer $p$)?

In Desmos, define m = 2. Then try one allowed value for n, such as n = 1, and choose simple integers like p = 0, q = 17 so that n*q = 17 is satisfied. Now define a = 2*p + n and create the four expressions a/m, a/n, 17/m, and 17/n. Observe which of these is a whole number. Then repeat with other possible values of n (17, -1, -17) that keep n*q = 17 true, and see which expression stays a whole number for every allowed choice of n. The one that is always a whole number is the correct choice.

Write out the product of $(mx+n)$ and $(x^2+px+q)$:

Now collect like terms:

This must equal $2x^3 + ax^2 + bx + 17$ for all $x$.

Compare the coefficients of $x^3$:

• From $mx^3 + (mp+n)x^2 + (mq+np)x + nq$, the $x^3$ coefficient is $m$.
• From $2x^3 + ax^2 + bx + 17$, the $x^3$ coefficient is $2$.

So $m = 2$.

This already lets you say something about expressions that involve $m$.

Now compare the constant terms (the terms without $x$):

• From the expanded form, the constant term is $nq$.
• From the original polynomial, the constant term is $17$.

So

Since $n$ and $q$ are integers and 17 is a prime number, the only possible integer factor pairs $(n,q)$ are:

• $(1,17)$
• $(17,1)$
• $(-1,-17)$
• $(-17,-1)$.

Match the coefficient of $x^2$ to understand how $a$ depends on $m$, $n$, and $p$:

• From the expanded form, the $x^2$ coefficient is $mp+n$.
• From the original polynomial, the $x^2$ coefficient is $a$.

So

Using $m = 2$, this becomes

This shows that $a$ depends on both $p$ and $n$ and is not fixed by $n$ alone.

We now know:

• $m = 2$, so any expression with $m$ is determined.
• $n$ must be one of $1, 17, -1, -17$.
• $a = 2p + n$, where $p$ is any integer.

Consider each option:

• $\dfrac{17}{m} = \dfrac{17}{2}$, which is not an integer.
• $\dfrac{a}{m} = \dfrac{2p+n}{2} = p + \dfrac{n}{2}$, which is not always an integer because $n$ can be odd.
• $\dfrac{a}{n} = \dfrac{2p+n}{n} = 2\cdot\dfrac{p}{n} + 1$, which is not guaranteed to be an integer since $p$ and $n$ are independent integers.
• For $\dfrac{17}{n}$: since $n$ must be a divisor of $17$ (one of $\pm1, \pm17$), $17/n$ is always an integer.

Therefore, the expression that must be an integer is $\dfrac{17}{n}$.