## Algebra 1: Common Core (15th Edition)

Published by Prentice Hall

# Chapter 8 - Polynomials and Factoring - 8-5 Factoring x2+bx+c - Practice and Problem-Solving Exercises - Page 515: 20

#### Answer

$q^2+3q-54=(q-6)(q+9)$

#### Work Step by Step

We are trying to fill in the blank in the equation $q^2+3q-54=(q-6)(q+\square)$. 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 $x^2+bx+c$, 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 $(q+\_)(q+\_)$. In the case of $q^2+3q-54$, we are looking for two numbers whose product is $-54$ and whose sum is $3$. The numbers $-6$ and $9$ meet these criteria, because $$9\times(-6)=-54\;\text{and}\;9+(-6)=3$$When we insert these numbers into the blanks, we arrive at the factors $(q-6)(q+3)$. Inserting factored form back into the original equation, we have the complete equation, which is $q^2+3q-54=(q-6)(q+9)$.

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