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: 39

Answer

$\left\{ -1,0,3 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ x^3-2x^2=3x ,$ in factored form. Then use the Zero Product Property by equating each factor to zero. Finally, solve each of the resulting equations. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} x^3-2x^2-3x=0 .\end{array} The $GCF$ of the constants of the terms $\{ 1,-2,-3 \}$ is $ 1 $ since it is the highest number that can divide all the given constants. The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ x^3,x^2,x \}$ is $ x .$ Hence, the entire expression has $GCF= x .$ Factoring the $GCF= x ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x(x^2-2x-3)=0 .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 1(-3)=-3 $ and the value of $b$ is $ -2 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 1,-3 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x(x^2+x-3x-3)=0 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} x[(x^2+x)-(3x+3)]=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x[x(x+1)-3(x+1)]=0 .\end{array} Factoring the $GCF= (x+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} x[(x+1)(x-3)]=0 \\\\ x(x+1)(x-3)=0 .\end{array} Equating each factor zero (Zero Product Property), then \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ x+1=0 \\\\\text{OR}\\\\ x-3=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ x+1=0 \\\\ x=-1 \\\\\text{OR}\\\\ x-3=0 \\\\ x=3 .\end{array} Hence, the solutions are $ \left\{ -1,0,3 \right\} .$
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