Algebra 1

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

Chapter 8 - Polynomials and Factoring - 8-4 Multiplying Special Cases - Practice and Problem-Solving Exercises: 42

Answer

$s^{2}+12st^{2}+36t^{4}$

Work Step by Step

$(s+6t^{2})^{2}$ Using the rule of $(a+b)^{2}$ = $a^{2}$+2ab+$b^{2}$ In this case, the a= s and b= $6t^{2}$ as substituting these gives us the original polynomial $a^{2}$+2ab+$b^{2}$ $s^{2}$+2(s)($6t^{2}$)+$(6t^{2})^{2}$ Thus, $s^{2}+12st^{2}+36t^{4}$
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