Answer
$22y^5+7y^4+8y^3-6y^2+6y-7$
Work Step by Step
Consider the given polynomial
$(12y^5+7y^4-3y^2+6y+7)-(-10y^5-8y^3+3y^2+14)$
Need to drop the parentheses and will have to change the sign of each term of the second polynomial because of the presence of the negative outside.
Thus,
$(12y^5+7y^4-3y^2+6y+7)-(-10y^5-8y^3+3y^2+14)=12y^5+7y^4-3y^2+6y+7+10y^5+8y^3-3y^2-14$
or, $=12y^5+7y^4-3y^2+10y^5+8y^3-3y^2+6y+7-14$
or, $=22y^5+7y^4+8y^3-6y^2+6y-7$
