From your comparison, look specifically at the equation involving the constant term $-35$. How are $k$ and $j$ connected to $-35$?

If $k$ and $j$ are integers and their product is $-35$, what can you say about how $k$ is related to $35$? Which answer choices involve a number being divided by something that must be a factor of $35$?

Express $b$ in terms of $j$ and $k$, and then rewrite $\dfrac{b}{6}$ and $\dfrac{b}{k}$. Do those forms guarantee an integer for every possible integer pair $(k, j)$ with product $-35$?

In Desmos, add variables k and j. Turn them into sliders and set them to take integer values. Choose a pair such that k*j = -35 (for example, select one factor pair of $-35$ like $5$ and $-7$).

In a new Desmos line, type b = 6*j + k to define the corresponding value of $b$ for your chosen $k$ and $j$. This matches the linear coefficient from expanding $(6x + k)(x + j)$.

On separate lines, type b/6, b/k, 35/6, and 35/k. Desmos will show decimal or integer values for each expression based on your current $k$ and $j$.

Change $k$ and $j$ to a different integer pair with product $-35$ (for example, swap the signs or use another factor pair). Watch how the four expressions change. Identify which expression stays as an integer for both choices of $(k, j)$—that expression corresponds to the correct answer choice.

Rewrite the given factored form and expand it:

This must equal $6x^2 + bx - 35$ for all $x$, so match coefficients of like terms:

• Coefficient of $x$: $b = 6j + k$
• Constant term: $kj = -35$

Here $k$ and $j$ are integers.

From the constant terms, we have

Because $k$ and $j$ are integers, this equation says that $k$ and $j$ are integer factor pairs of $-35$.

That means:

• $k$ must be an integer divisor (factor) of $35$.
• And $j = \dfrac{-35}{k}$ for each such integer $k$.

Now use what we know about $k$ and $j$ to check each option.

Recall:

• $b = 6j + k$
• $j = \dfrac{-35}{k}$

1. Option A: $\dfrac{b}{6}$

   Substitute $b = 6j + k$:

Since $k$ does not have to be a multiple of $6$, $\dfrac{k}{6}$ is not always an integer, so $\dfrac{b}{6}$ is not guaranteed to be an integer.

2. Option B: $\dfrac{b}{k}$

There is no reason $k$ must divide $6j$, so $\dfrac{6j}{k}$ is not always an integer. Thus $\dfrac{b}{k}$ is not guaranteed to be an integer.

3. Option C: $\dfrac{35}{6}$

   This value does not depend on $k$ or $j$ at all. It is a fixed number, and $35$ is not divisible by $6$, so $\dfrac{35}{6}$ is not an integer.

4. Option D: $\dfrac{35}{k}$

   From step 2, $k$ is always an integer factor of $35$, so $k$ always divides $35$. That means $\dfrac{35}{k}$ is always an integer.

Therefore, the expression that must be an integer is $\dfrac{35}{k}$.