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: 46



Work Step by Step

The two numbers whose product is $ac= 1(54)=54 $ and whose sum is $b= -15 $ are $\{ -6,-9 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ x^2-15x+54 $, is \begin{array}{l}\require{cancel} x^2-6x-9x+54 \\\\= (x^2-6x)-(9x-54) \\\\= x(x-6)-9(x-6) \\\\= (x-6)(x-9) .\end{array}
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