Notice that every option looks like $2x^2 + bx + 15a$ for some number $b$. Try substituting $x=3a$ into this general form instead of doing each option from scratch.

After substituting $x=3a$, set the expression equal to $0$ and solve for $a$ in terms of $b$. Then plug in each option’s $b$ and see which one gives a positive integer value for $a$.

In Desmos, type a=1. Desmos will prompt you to add a slider for a. Make sure the slider step is set to 1 and that it covers several positive integers (for example, from 1 to 10).

For each option, substitute $x=3a$ and type the resulting expression as a function of $a$. For example, for choice A type f_A(a) = 2(3a)^2 + 17(3a) + 15a, and similarly define expressions for the other choices with their respective $x$-coefficients.

Move the a slider through positive integers and watch the values of each expression. The correct choice is the one whose substituted expression becomes exactly 0 for some positive integer value of a.

If $x-3a$ is a factor of a polynomial, then $x=3a$ is a root of that polynomial.

So for any expression that has $x-3a$ as a factor, when we substitute $x=3a$ into the expression, the result must be $0$.

All four answer choices have the form

where $b$ is the coefficient of $x$ (in the options, $b$ is $17$, $-31$, $-29$, or $-19$).

Substitute $x=3a$ into this general form:

For $x-3a$ to be a factor, we need this expression to equal $0$ when $a$ is a positive integer.

Set the substituted expression equal to $0$:

Since $a$ must be a positive integer, $a \neq 0$, so we require

Solve for $a$:

Now plug in each option’s $b$ value:

• Choice A: $b = 17$ gives $a = \dfrac{-17 - 5}{6} = -\dfrac{11}{3}$ (not positive).
• Choice B: $b = -31$ gives $a = \dfrac{31 - 5}{6} = \dfrac{13}{3}$ (not an integer).
• Choice C: $b = -29$ gives $a = \dfrac{29 - 5}{6} = 4$ (a positive integer).
• Choice D: $b = -19$ gives $a = \dfrac{19 - 5}{6} = \dfrac{7}{3}$ (not an integer).

Only the expression $2x^2 - 29x + 15a$ (choice C) allows a positive integer $a$ for which $x-3a$ is a factor (specifically $a=4$).