Question 618338
I'm assuming that the last term in the numerator is just 46, not 46b...
{{{(30b^3+13b^2+44b+46)/(6b+5)}}}
Using long division:
<pre>
       5b^2 + (-2b) +     9
     ___________________________
6b+5/ 30b^3 + 13b^2 +   44b + 46
      30b^2 + 25b^2
      =============
             -12b^2 +   44b
             -12b^2 + (-10b)
             ===============
                        54b + 46
                        54b + 45
                        ========
                               1
</pre>
So 
{{{(30b^3+13b^2+44b+46b)/(6b+5) = 5b^2 + (-2b) + 9 + 1/(6b+5)}}}<br>
P.S. If the last term of the numerator was 45, then there would be no remainder and the answer would just be {{{5b^2 + (-2b) +     9}}}