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 303: 40



Work Step by Step

Let $z=x^3$. Then the given expression, $ x^6-4x^3-12 $, is equivalent to $ z^2-4z-12 $. The two numbers whose product is $ac= 1(-12)=-12 $ and whose sum is $b= -4 $ are $\{ -6,2 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ z^2-4z-12 $, is \begin{array}{l}\require{cancel} z^2-6z+2z-12 \\\\= (z^2-6z)+(2z-12) \\\\= z(z-6)+2(z-6) \\\\= (z-6)(z+2) .\end{array} Since $z=x^3$, then, \begin{array}{l} (z-6)(z+2) \\\\= (x^3-6)(x^3+2) .\end{array}
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