## Precalculus (6th Edition)

$12y^{2}+4$
First distribute the two located in front of the first set of parenthesis so that you are left with: $$(24y^{2}-6y+12)-4(3y^{2}-4y+2)$$Then distribute the $-4$ located in front of the second set of parenthesis so that you are left with: $$24y^{2}-16y+12-12y^{2}+16y-8$$ Then combine like terms (squared with squared, etc.) to get the final answer: $$12y^{2}+4$$