How do you divide `(3a^3+17a^2+12a-5)/(a+5)`?

First, let's split the top polynomial up into "multiples" of the bottom polynomial. To see what I mean, let's first try to take care of the `3a^3` term.

We can see that `3a^2(a+5) = 3a^3+15a^2`. So, let's separate this from our polynomial:

`3a^3 + 17a^2 + 12a - 5`

`(3a^3+15a^2) + 2a^2+12a-5`

`3a^2(a+5) + 2a^2+12a-5`

See how this "gets rid of" the `3a^3` term? Let's do the same for the `2a^2` term.

We can see that `2a(a+5) = 2a^2 + 10a`.

`3a^2(a+5) + (2a^2+10a) + 2a-5`

`3a^2(a+5) + 2a(a+5)+ 2a-5`

The next highest term to deal with is the `2a` term.

We can see that `2(a+5) = 2a+10`.

`3a^2(a+5) + 2a(a+5)+ (2a+10) -15`

`3a^2(a+5) + 2a(a+5)+ 2(a+5) - 15`

We can't do anything about the `15`, since its degree is smaller than the degree of `a+5`.

Finally, let's divide everything by `(a+5)`. This is why we wrote everything in terms of `a+5`, so we can just cancel out the top and the bottom of the fraction at this step.

`(3a^3+17a^2+12a-5)/(a+5)`

`= (3a^2(a+5) + 2a(a+5)+ 2(a+5) - 15)/(a+5)`

`= 3a^2+2a+2 - 15/(a+5)`

Final Answer