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 - Page 515: 23

Answer

$y^2+8y-9=(y+9)(y-1)$

Work Step by Step

We are trying to fill in the blank in the equation $y^2+8y-9=(y+9)(y-\square)$. 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 $y^2+by+c$, 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 $(y+\_)(y+\_)$. In the case of $y^2+8y-9$, we are looking for two numbers whose product is $-9$ and whose sum is $8$. The numbers $9$ and $-1$ meet these criteria because $$9\times(-1)=-9\;\text{and}\;9+(-1)=8$$When we insert these numbers into the blanks, we arrive at the factors: $(y+9)(y-1)$.
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