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 - Practice and Problem-Solving Exercises: 56

Answer

3$(6n^{3}+7)(6n^{3}-7)$

Work Step by Step

3($36n^{6}$ - 49) We use the formula for the difference of squares to apply to this question. The difference of squares formula is: $(a-b) (a+b) = a^{2} - b^{2}$ = 3($36n^{6}$ - 49) *** Take the square root of $36n^{6}$ which is $6n^{3}$. Becuase $6n^{3}$ × $6n^{3}$ = $36n^{6}$ *** Take the square root of 49 which is 7. Becuase 7 × 7= 49 = $(6n^{3})^{2}−7^{2}$ In the given formula let $6n^{3}$ represents a and 7 represents b. 3($(6n^{3})^{2}−7^{2}$) = 3$(6n^{3}+7)(6n^{3}-7)$
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