## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Section 5.5 - Solving Equations by the Zero-Factor Property - 5.5 Exercises: 45

#### Answer

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

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $x^3-6x^2-9x+54=0 ,$ in factored form. Then use the Zero Product Property by equating each factor to zero. Finally, solve each of the resulting equations. $\bf{\text{Solution Details:}}$ Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (x^3-6x^2)-(9x-54)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x^2(x-6)-9(x-6)=0 .\end{array} Factoring the $GCF= (x-6)$ of the entire expression above results to \begin{array}{l}\require{cancel} (x-6)(x^2-9)=0 .\end{array} The expressions $x^2$ and $9$ are both perfect squares and are separated by a minus sign. Hence, $x^2-9 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x-6)(x+3)(x-3)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} x-6=0 \\\\\text{OR}\\\\ x+3=0 \\\\\text{OR}\\\\ x-3=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-6=0 \\\\ x=6 \\\\\text{OR}\\\\ x+3=0 \\\\ x=-3 \\\\\text{OR}\\\\ x-3=0 \\\\ x=3 .\end{array} Hence, the solutions are $x=\left\{ -3,3,6 \right\} .$

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