Dividing a college algebra fraction?

Question

27z+23z^2+6z^3

______________

2z+3

Answer 1

= (27z + 23z² + 6z³)/(2z + 3)

= 3z² + 7z + 3 remainder - 9

Answer: 3z² + 7z + 3 & - 9/(2z + 3)

Checking (going back to 27z + 21z² + 6z³):

= ([2z + 3][3z²+ 7z + 3]) + (- 9)

= (6z³ + 14z² + 6z + 9z² + 21z + 9) - 9

= 6z³ + 14z² + 9z² + 6z + 21z + 9 - 9

= 6z³ + 23z² + 27z + 0

= 6z³ + 23z² + 27z or

= 27z + 23z² + 6z³

Answer 2

        3Z^2+7Z

        ----------------------------------

2z+3 I 6Z^3 + 23Z^2 + 27Z

          6Z^3 +  9Z^ 2

        ----------------------------

                    14Z^2 +27Z

                     14Z^2+ 21Z

         -----------------------------------

                               6Z

Answer 3

you just do regular long division

put 2z+3 on the outside of the house and 6z^3 + 23z^2 +27z on the inside.  So fir your first number that goes on the roof it would be 3z^2 and then you would minus 6z^3 + 9z^2  from the numerator, and keep continuing long division.