Question 224799
To do this problem, you have to use long division.
Set up the equation with the {{{(35b^3+18b^2+37b+39)}}} under the division symbol, and the {{{5b+4}}} on the outside.
The first one to look at is {{{35b^3/5b}}}
The answer to this is {{{7b^2}}}.
This we have to multiply back with {{{5b+4}}} to determine what is going to be subtracted from our equation:
So, we end up with:
{{{+35b^3+18b^2}}} which is from our initial equation
{{{-35b^3+28b^2}}} which is from multiplying our answer and the {{{5b+4}}}
which is {{{0-10b^2}}}.
We bring down the {{{37b}}} and have a new equation to evaluate: {{{-10b^2+37b}}}.
We divide the first term by the {{{5b}}} again to give us: {{{-2b}}}.
Multiplying this through gives us:
{{{-10b^2+37b}}} which is from our new equation
{{{+10b^2+8b}}} which is from multiplying our answer and the {{{5b+4}}}
which is {{{0+45b}}}.
We bring down the 3 and have a new equation to evaluate: {{{45b+3}}}
Divide the first term by {{{5b}}} leaves us with 9.
Multiplying through gives us:
{{{45b+3}}} which is from our new equation
{{{45b+36}}} which is from multiplying our answer and the {{{5b+4}}}
This leaves 39, which is our remainder.

The entire answer after working all this out is:
{{{7b^2-2b+9+39/(5b+4)}}}
The {{{39/(5b+4)}}} is our remainder.