Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 8 - Polynomials and Factoring - 8-7 Factoring Special Cases - Lesson Check - Page 514: 1



Work Step by Step

$y^{2}$-16y+64 Using the rule of $a^{2}$-2ab+$b^{2}$= $(a-b)^{2}$ In this case, the a= y and b= 8 as substituting these gives us the original polynomial $a^{2}$-2ab+$b^{2}$ $y^{2}$-2(y)(8)+$8^{2}$ $y^{2}$-16y+64 Thus: $y^{2}$-16y+64 = $(y-8)^{2}$ = (y-8)(y-8)
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