SAT Question e51bf5b1: Equivalent Expressions answer and explanation with Desmos solution

Choice D is correct. Since each choice has a term of $3x^2$, which can be written as $\left(3x\right)\left(x\right)$, and each choice has a term of $14b$, which can be written as $\left(7\right)\left(2b\right)$, the expression that has a factor of $x+2b$, where $b$ is a positive integer constant, can be represented as $\left(3x+7\right)\left(x+2b\right)$. Using the distributive property of multiplication, this expression is equivalent to $3x\left(x+2b\right)+7\left(x+2b\right)$, or $3x^2+6xb+7x+14b$. Combining the x-terms in this expression yields $3x^2+\left(7+6b\right)x+14b$. It follows that the coefficient of the x-term is equal to $7+6b$. Thus, from the given choices, $7+6b$ must be equal to $7$, $28$, $42$, or $49$. Therefore, $6b$ must be equal to $0$, $21$, $35$, or $42$, respectively, and $b$ must be equal to $\frac{0}{6}$, $\frac{21}{6}$, $\frac{35}{6}$, or $\frac{42}{6}$, respectively. Of these four values of $b$, only $\frac{42}{6}$, or $7$, is a positive integer. It follows that $7+6b$ must be equal to $49$ because this is the only choice for which the value of $b$ is a positive integer constant. Therefore, the expression that has a factor of $x+2b$ is $3x^2+49x+14b$.