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: 36

Answer

$(x^2-5)(x^2+4)$

Work Step by Step

Let $z=x^2$. Then the given expression, $ x^4-x^2-20 $, is equivalent to $ z^2-z-20 $. The two numbers whose product is $ac= 1(-20)=-20 $ and whose sum is $b= -1 $ are $\{ 4,-5 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ z^2-z-20 $, is \begin{array}{l}\require{cancel} z^2-5z+4z-20 \\\\= (z^2-5z)+(4z-20) \\\\= z(z-5)+4(z-5) \\\\= (z-5)(z+4) .\end{array} Since $z=x^2$, then the factored form of the original expression is $ (x^2-5)(x^2+4) $.
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