Question 18·Hard·Advanced Math

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Which of the following polynomial expressions is divisible by $x + 3a$ , where $a$ is a positive integer constant?

For polynomial divisibility by a linear factor like $x+3a$ , immediately think of the Remainder Theorem: a polynomial is divisible by $x-c$ if substituting $x=c$ makes the polynomial equal zero. Instead of doing long division, plug $x=-3a$ into each option, simplify carefully by grouping $a^3$ , $a^2$ , and $a$ terms, and choose the expression that evaluates to 0; this is typically faster and less error-prone on the SAT.

Hints

Hint 1
Use the meaning of “divisible by” for polynomials
If a polynomial is divisible by $x+3a$ , what does that say about the value of the polynomial when $x=-3a$ ? Think about the Remainder Theorem or about what it means for $x=-3a$ to be a root.
Hint 2
Apply the Remainder Theorem
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.
Hint 3
Focus on substitution and combining like terms
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.

Desmos guide

Step 1
Set a value for a
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).
Step 2
Enter each polynomial as a function
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.
Step 3
Evaluate each function at x = -3a
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 $x+3a$ .

Step-by-step

1
Step 1
Translate “divisible by $x + 3a$ ” into an equation
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 $0$ .
2
Step 2
Set up the substitution for each option
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 $x=-3a$ .
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3
Step 3
Evaluate each option at $x=-3a$ and pick the one that gives 0
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 $x + 3a$ .

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Question 18·Hard·Advanced Math

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