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.
