Question 248231
#2 one way is to  start by setting it up like long division.
the first step
y+3 |into| 6y4 + 15y3 goes 6y3 times
6y^3(y+3)=6y^4+18y^3
so we subtract from 6y4 + 15y3 and get -3y^3 remember that 15-18=-3
how many times does y go into -3y^3 answer -3y^2
we don't have any y^2 in our term
so multiply out (y+3)(-3y^2) and get -3y^3-9y^2
so next we subtract 9y^2 from zero and get 9y^2
how many times does y go into 9y^2
9y of course
(y+3)*9y=9y^2+27y
but we have 28y
so we get y as the remainder
now we have y+6 
y+3 goes into y+6 once with a remainder of 3
so we have (6y^3-3y^2+9y+1)(y+3) +1