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c) First, we set them equal to each other to see for any value x it is true.

By setting each expresstion equal to each other, we have $\frac{6x^3+9x^2+4x+7}{2x+3}=3x^2+2+\frac{1}{2x+3}$.multiplying by 2x+3 on each side gets us $6x^3 + 9x^2+4x+7=3x^2(2x+3)+2(2x+3)+1.$Simlifying the right hand side leaves us with $6x^3+9x^2+4x+7=6x^3+9x^2+4x+6+1$. As you can see, we end up with$0=0,$witch is a true equation, and therefore any value of x works. But, a fraction can not be diuvvide by zero, and if x=-2/3, plugged into 2x+3, makes you didvide by zero. so, All values of x work except for -3/2.