Question 311190
{{{(y^3+8y^2+14y+12)/(y+6)}}}
First factor {{{y^2}}}
{{{y2(y+6)=y^3+6y^2}}}
Subtracted from the original polynomial gives a remainder of,
{{{(y^3+8y^2+14y+12)-(y^3+6y^2)=2y^2+14y+12}}}
Second factor {{{2y}}}
{{{2y(y+6)=2y^2+12y}}}
Subtracted from the previous remainder gives a new remainder of,
{{{(2y^2+14y+12)-(2y^2+12y)=2y+12}}}
Third factor: {{{2}}}
{{{2(y+6)=2y+12}}}
Subtracted from the previous remainder gives a remainder of 0.
{{{2y+12-(2y+12)=0}}}

Gather up all of the factors.
{{{(y^3+8y^2+14y+12)/(y+6)=y^2+2y+2}}}