sat-dog

3 2 Choice D is correct. Since each choice has a term of x , which can be written as 14 7 2 ^3xh^xh, and each choice has a term of b, which can be written as ^ h^ bh, the + expression that has a factor of x 2b, where b is a positive integer constant, can 3 +7 +2 be represented as ^ x h^x bh. Using the distributive property of 3 +2 +7 +2 multiplication, this expression is equivalent to x^x bh ^x bh, or 3 2+6 +7 +14 x xb x b. Combining the x-terms in this expression yields 3 2+ 7+6 +14 x ^ bhx b. It follows that the coefficient of the x-term is equal to 7+6 7+6 28 42 49 b. Thus, from the given choices, b must be equal to 7, , , or . 21 35 42 Therefore, 6b must be equal to 0, , , or , respectively, and b must be equal 0 21 35 42 42 to 6, 6, 6, or 6, respectively. Of these four values of b, only 6, or 7, is a 7+6 49 positive integer. It follows that b must be equal to because this is the only choice for which the value of b is a positive integer constant. Therefore, the + 3 2+49 +14 expression that has a factor of x 2b is x x b. + 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 21 + x 2b, then the value of b is 6, which isn’t an integer. Choice C is incorrect. If 35 + this expression has a factor of x 2b, then the value of b is 6, which isn’t an integer.