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: 29


$x=\left\{ -6,\dfrac{1}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ (2x+1)(x-3)=6x+3 ,$ in $ax^2+bx+c=0$ form. Then, express the equation in factored form. Next, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation. $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 2x(x)+2x(-3)+1(x)+1(-3)=6x+3 \\\\ 2x^2-6x+x-3=6x+3 \\\\ 2x^2+(-6x+x-6x)+(-3-3)=0 \\\\ 2x^2-11x-6=0 .\end{array} To factor the equation 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 $ 2(-6)=-12 $ and the value of $b$ is $ -11 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-12\}, \{2,-6\}, \{3,-4\} \\ \{-1,12\}, \{-2,6\}, \{-3,4\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -1,12 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2x^2-x+12x-6=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} (2x^2-x)+(12x-6)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(2x-1)+6(2x-1)=0 .\end{array} Factoring the $GCF= (2x-1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (2x-1)(x+6)=0 .\end{array} Equating each of the factors in the expression above to zero (Zero-Factor Theorem), then, \begin{array}{l}\require{cancel} 2x-1=0 \text{ OR } x+6=0 .\end{array} Solving each of the equations above results to \begin{array}{l}\require{cancel} 2x=1 \\\\ x=\dfrac{1}{2} \\\\\text{ OR } x+6=0 \\\\ x=-6 .\end{array} Hence, the solutions are $ x=\left\{ -6,\dfrac{1}{2} \right\} .$
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