OG详解-OG10 数学2 Q25

Choice D is correct. Since each choice has a term of 3x² , which can be written as (3x)(x), and each choice has a term of 14b, which can be written as (7)(2b), the expression that has a factor of x+2b, where b is a positive integer constant, can be represented as (3x+7)(x+2b). Using the distributive property of multiplication, this expression is equivalent to 3x(x+2b)+7(x+2b), or 3x² +6xb+7x+14b. Combining the x-terms in this expression yields 3x² +(7+6b)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 06​​ , 21​6​ , 35​6​, or 42​6​, respectively. Of these four values of b, only 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 3x² +49x+ 14b.
Choice A is incorrect. If this expression has a factor of x+2b, then the value of b is 0, which isn’t positive. Choice B is incorrect. If this expression has a factor of x+2b, then the value of b is 21​6​, which isn’t an integer. Choice C is incorrect. If this expression has a factor of x+2b, then the value of b is 35​6​, which isn’t an integer.