Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.3 Polynomials - R.3 Exercises - Page 33: 51

Answer

$12y^{2}+4$

Work Step by Step

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: $$12y^{2}+4$$
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