Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 8 - Polynomials and Factoring - 8-5 Factoring x2+bx+c - Practice and Problem-Solving Exercises: 21

Answer

$z^2-2z-48=(z-8)(z+6)$

Work Step by Step

We are trying to fill in the blank in the equation $z^2-2z-48=(z-8)(z+\square)$. In order to do so, we will factor the trinomial on the left side of the equation to determine the second factor. To factor a trinomial in the form $z^2+bz+c$, we must find two numbers whose product is $c$ and whose sum is $b$. We then insert these two numbers into the blanks of the factors $(z+\_)(z+\_)$. In the case of $z^2-2z-48$, we are looking for two numbers whose product is $-48$ and whose sum is $-2$. The numbers $-8$ and $6$ meet these criteria, because $$6\times(-8)=-48\;\text{and}\;6+(-8)=-2$$When we insert these numbers into the blanks, we arrive at the factors: $(z-8)(z+6)$.
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