Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.5 - Solving Equations by the Zero-Factor Property - 5.5 Exercises - Page 355: 9


$x=\left\{ -6,-3 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ x^2+9x+18=0 ,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation. $\bf{\text{Solution Details:}}$ The factored form of the equation above is \begin{array}{l}\require{cancel} (x+3)(x+6)=0 .\end{array} Equating each factor to zero (Zero-Factor Property), then \begin{array}{l}\require{cancel} x+3=0 \text{ OR } x+6=0 .\end{array} Using the properties of equality to solve each of the equation above results to \begin{array}{l}\require{cancel} x+3=0 \\\\ x=-3 \\\\\text{ OR }\\\\ x+6=0 \\\\ x=-6 .\end{array} Hence, the solutions are $ x=\left\{ -6,-3 \right\} .$ $\bf{\text{Supplementary Solution:}}$ To factor the expression, $ x^2+9x+18 ,$ find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ Then, express the factored form as $(x+m_1)(x+m_2).$ In the given expression, the value of $c$ is $ 18 $ and the value of $b$ is $ 9 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,18 \}, \{ 2,9 \}, \{ 3,6 \}, \{ -1,-18 \}, \{ -2,-9 \}, \{ -3,-6 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 3,6 \}.$ Hence, the factored form of the expression above is \begin{array}{l}\require{cancel} (x+3)(x+6) .\end{array}
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