Question 135106
First of all, you have an error in the statement of the problem:

You show: {{{(2x-3)(red(3x-5))+ r = 6x^2 + x + 5}}}


And it should be {{{(2x-3)(green(3x+5))+ r = 6x^2 + x + 5}}}


Use polynomial long division or synthetic division to divide {{{6x^2 + x + 5}}}
 by either {{{2x-3}}} or {{{3x+5}}}.


{{{2x}}} goes into {{{6x^2}}} {{{3x}}} times.  So {{{3x}}} is the first term of your quotient.


{{{3x}}} times {{{2x-3}}} is {{{6x^2-9x}}}


Subtract {{{6x^2-9x}}} from {{{6x^2+x}}} resulting in {{{10x}}} (remember, change the sign and add)


Bring down the {{{5}}}


{{{2x}}} goes into {{{10x}}} {{{5}}} times.  So {{{5}}} is the second term of your quotient.


{{{5}}} times {{{2x-3}}} is {{{10x-15}}}


Subtract {{{10x-15}}} from {{{10x+5}}} resulting in {{{20}}} which is your remainder.


So:  {{{(2x-3)(3x+5)+20=6x^2+x+5}}}


Check the answer:
{{{(2x-3)(3x+5)+20=6x^2+x-15+20=6x^2+x+5}}}  Checks.