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: 39

Answer

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

Work Step by Step

Let $z=x^3$. Then the given expression, $ x^6-7x^3+12 $, is equivalent to $ z^2-7z+12 $. The two numbers whose product is $ac= 1(12)=12 $ and whose sum is $b= -7 $ are $\{ -3,-4 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ z^2-7z+12 $, is \begin{array}{l}\require{cancel} z^2-3z-4z+12 \\\\= (z^2-3z)-(4z-12) \\\\= z(z-3)-4(z-3) \\\\= (z-3)(z-4) .\end{array} Since $z=x^3$, then, \begin{array}{l} (z-3)(z-4) \\\\= (x^3-3)(x^3-4) .\end{array}
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