Solution:
In f(x), factoring out the greatest common factor, 2x, yields f(x)=2x(x2+3x+2).It is given that g(x)=x2+3x+2,so using substitution,f(x) can be rewritten as f(x)=2x·g(x).In the equation p(x)=f(x)+3g(x), substituting 2x·g(x)forf(x) yields p(x)=2x·g(x)+3·g(x). In p(x), factoring out the greatest common factor,g(x), yields p(x)=(g(x))(2x+3). Because 2x+3 is a factor of p(x), it follows that p(x) is divisible by 2x+3.
Choices A, C, and D are incorrect because 2x+3 is not a factor of the polynomials h(x),r(x), or s(x).Using the substitution f(x)=2x·g(x), and factoring further, h(x),r(x), and s(x) can be rewritten as follows:
h(x)=(x+1)(x+2)(2x+1)
r(x)=(x+1)(x+2)(4+3)
s(x)=2(x+1)(x+2)(3x+1)Because 2x+3 is not a factor of h(x),r(x), or s(x),it follows that h(x), r(x), and s(x) are not divisible by 2x+3.
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