Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.6 - Factoring Trinomials - Exercise Set - Page 304: 82

Answer

$(x^3-4)(3x^3-2)$

Work Step by Step

The two numbers whose product is $ac= 3(8)=24 $ and whose sum is $b= -14 $ are $\{ -12,-2 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ 3x^6-14x^3+8 $, is \begin{array}{l}\require{cancel} 3x^6-12x^3-2x^3+8 \\\\= (3x^6-12x^3)-(2x^3-8) \\\\= 3x^3(x^3-4)-2(x^3-4) \\\\= (x^3-4)(3x^3-2) .\end{array}
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