From matching coefficients, what must $p$ equal? What equation do you get from the constant term $-28$ in terms of $q$ and $r$? Remember $p$, $q$, and $r$ are all integers.

Use your equations (especially $p=9$, $qr=-28$, and $m=9r+q$) to rewrite each answer choice in terms of $p$, $q$, and $r$. For A, B, and C, can you find specific integer values of $q$ and $r$ (with $qr=-28$) that make those expressions not integers?

Because $q$ and $r$ are integers with product $-28$, each is a divisor of $-28$. Use that idea to see which fraction in the choices is always equal to an integer like $p$, $q$, or $r$ (possibly up to a sign).

In Desmos, add definitions that mirror the algebra: type p = 9. Then choose an integer value for q that is a divisor of 28 (for example, use a slider q from -28 to 28 with step 1, but only stop on values that divide 28 evenly). Define r = -28/q. Finally define m = 9*r + q.

On new lines, enter the four expressions from the choices: m/p, m/q, 28/p, and 28/q. Desmos will display numeric values for each expression based on your current q (and resulting r and m).

Move the q slider through different integer divisors of 28 (such as ±1, ±2, ±4, ±7, ±14, ±28). Each time, check which of the four expressions is always an integer while the others sometimes become non-integers. The expression that consistently remains an integer for all valid q values corresponds to the correct choice.

Rewrite the given equality in expanded form.

Expand $(px+q)(x+r)$:

This must equal the original quadratic:

So the coefficients of like powers of $x$ must match:

• Coefficient of $x^2$: $p = 9$.
• Coefficient of $x$: $pr + q = m$.
• Constant term: $qr = -28$.

From Step 1, we have three key equations, with $p,q,r$ all integers:

• $p = 9$ (so $p$ is fixed).
• $qr = -28$.
• $m = pr + q = 9r + q$.

Because $q$ and $r$ are integers and $qr = -28$, $q$ and $r$ are integer factors of $-28$. That means there are many possible integer pairs $(q,r)$ satisfying $qr=-28$, and for each such pair we get a (possibly different) $m$.

Now analyze some answer choices using $p=9$ and $m=9r+q$.

Choice A: $\dfrac{m}{p}$

Using $p=9$ and $m=9r+q$:

This is an integer only when $q$ is a multiple of 9, which is not always true for factors of $-28$.

Example: Take $q=4$, $r=-7$ (since $4\cdot(-7)=-28$). Then $m = 9(-7)+4 = -59$, so

which is not an integer. So $\dfrac{m}{p}$ does not have to be an integer.

Choice B: $\dfrac{m}{q}$

Again use $m=9r+q$:

For this to always be an integer, $r/q$ would need to always be an integer, but that is not guaranteed.

Using the same example $q=4$, $r=-7$ gives $m=-59$ and

which is not an integer. So $\dfrac{m}{q}$ does not have to be an integer.

Now consider the remaining choices using the relationships above.

Choice C: $\dfrac{28}{p}$

Since $p=9$,

which is not an integer. So $\dfrac{28}{p}$ is not guaranteed to be an integer.

For the last expression, use the constant-term equation $qr=-28$.

From $qr = -28$ and $q \ne 0$, divide both sides by $q$:

Since $r$ is an integer, $\frac{-28}{q}$ must be an integer. That means

is also always an integer.

Therefore, the expression that must be an integer is

which corresponds to choice D.