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 356: 37


$x=\left\{ -\dfrac{1}{2},0,5 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ 2x^3-9x^2-5x=0 ,$ in factored form. Next, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation. $\bf{\text{Solution Details:}}$ The $GCF$ of the constants of the terms $\{ 2,-9,-5 \}$ is $ 1 .$ 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 \left( 2x^2-9x-5 \right)=0 .\end{array} To factor the trinomial expression above, find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ In the expression above the value of $ac$ is $x \left( 2x^2-9x-5 \right)=0 2(-5)=-10 $ and the value of $b$ is $ -9 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-10\}, \{2,-5\}, \\ \{-1,10\}, \{-2,5\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 1,-10 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x \left( 2x^2+1x-10x-5 \right)=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 [(2x^2+1x)-(10x+5)]=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x [x(2x+1)-5(2x+1)]=0 .\end{array} Factoring the $GCF= (2x+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} x [(2x+1)(x-5)]=0 \\\\ x(2x+1)(x-5)=0 .\end{array} Equating each of the factors in the expression above to zero (Zero-Factor Theorem), then, \begin{array}{l}\require{cancel} x=0 \text{ OR } 2x+1=0 \text{ OR } x-5=0 .\end{array} Solving each of the equations above results to \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ 2x+1=0 \\\\ 2x=-1 \\\\ x=-\dfrac{1}{2} \\\\\text{ OR }\\\\ x-5=0 \\\\ x=5 .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{1}{2},0,5 \right\} .$
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