Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 6 - Section 6.5 - Factoring by Special Products - Exercise Set: 8

Answer

$36a^{2}$ - 12ab + $b^{2}$ This is a perfect square trinomial. $36a^{2}$ - 12ab + $b^{2}$ = $(6a - b)^{2}$

Work Step by Step

$36a^{2}$ - 12ab + $b^{2}$ Notice that both the first and last terms are perfect squares: $36a^{2}$ = 6a $\times$ 6a or = -6a $\times$ -6a and $b^{2}$ = b $\times$ b or = -b $\times$ -b. Also the middle term is -12ab = $2$ $\times$ $-6a$ $\times$ b. Thus this is a perfect square trinomial: $36a^{2}$ - 12ab + $b^{2}$ = $(6a)^{2}$ + 2(6a)(-b) + $(-b)^{2}$ = $(6a - b)^{2}$.
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