For a polynomial $P(x)$, the remainder when dividing by $x-c$ is $P(c)$. Use this idea with $c=-3a$ and check what must be true if $x+3a$ is a factor.

Substitute $x=-3a$ into each option. Be careful with exponents on $-3a$, and then combine the $a^3$, $a^2$, and $a$ terms separately to see which expression becomes 0.

In Desmos, type a = 1 to create a slider for the positive integer constant $a$ (you can later change it to 2, 3, etc. to double-check).

For each option, define a function in Desmos, for example: f(x) = 2x^3 + (6a + 5)x^2 + (15a + 6)x + 18a for one choice, and similarly define functions for the other choices with their exact coefficients.

On new lines, type f(-3a), g(-3a), etc. for each function you defined. Look at the numerical outputs; the function whose value at $x=-3a$ is exactly 0 (for your positive integer value of $a$) corresponds to the polynomial that is divisible by $x+3a$.

If a polynomial $P(x)$ is divisible by $x+3a$, then $x+3a$ is a factor of $P(x)$.

That means $x=-3a$ must be a root of $P(x)$, so by the Remainder Theorem:

• $P(x)$ is divisible by $x+3a$ if and only if $P(-3a)=0$.

So the plan is: for each answer choice, plug in $x=-3a$ and see whether the result equals $0$.

Take a generic expression of the form

• $2x^3 + (6a+5)x^2 + (\text{something})x + (\text{something with }a)$.

When you substitute $x=-3a$:

• $2x^3$ becomes $2(-3a)^3 = -54a^3$.
• $(6a+5)x^2$ becomes $(6a+5)(-3a)^2 = (6a+5)\cdot 9a^2 = 54a^3 + 45a^2$.

These first two terms are the same in every option and they always add to $0a^3 + 45a^2$ (because $-54a^3 + 54a^3 = 0$). The difference between the options comes from:

• the coefficient of $x$ and
• the constant term (the part without $x$).

So you only need to be careful with what happens to the $x$-term and constant term when $x=-3a$.

Now plug $x=-3a$ into each full polynomial.

Option A: $2x^3 + (6a + 5)x^2 + (15a + 6)x + 18a$

• $2(-3a)^3 = -54a^3$
• $(6a+5)(-3a)^2 = 54a^3 + 45a^2$
• $(15a+6)(-3a) = -45a^2 - 18a$
• $18a$ stays $18a$

Add like terms:

• Cubic: $-54a^3 + 54a^3 = 0$
• Quadratic: $45a^2 - 45a^2 = 0$
• Linear: $-18a + 18a = 0$

So for option A, $P(-3a)=0$.

Option B: $2x^3 + (6a + 5)x^2 + (12a + 6)x + 18a$

• First two terms still give $0a^3 + 45a^2$.
• $(12a+6)(-3a) = -36a^2 - 18a$
• Constant term: $18a$

Combine:

• Quadratic: $45a^2 - 36a^2 = 9a^2$ (not $0$ if $a>0$)
• Linear: $-18a + 18a = 0$

So $P(-3a) = 9a^2 \neq 0$.

Option C: $2x^3 + (6a + 5)x^2 + (15a + 4)x + 18a$

• First two terms: $0a^3 + 45a^2$.
• $(15a+4)(-3a) = -45a^2 - 12a$
• Constant term: $18a$

Combine:

• Quadratic: $45a^2 - 45a^2 = 0$
• Linear: $-12a + 18a = 6a \neq 0$.

Option D: $2x^3 + (6a + 5)x^2 + (15a + 6)x + 12a$

• First two terms: $0a^3 + 45a^2$.
• $(15a+6)(-3a) = -45a^2 - 18a$
• Constant term: $12a$

Combine:

• Quadratic: $45a^2 - 45a^2 = 0$
• Linear: $-18a + 12a = -6a \neq 0$.

Only option A gives $P(-3a)=0$, so the expression in option A, $2x^3 + (6a + 5)x^2 + (15a + 6)x + 18a$, is the one that is divisible by $x + 3a$.