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: 34


$x=\left\{ -\dfrac{4}{3},5 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ (3x+1)(x-3)=2+3(x+5) ,$ 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} 3x(x)+3x(-3)+1(x)+1(-3)=2+3(x+5) \\\\ 3x^2-9x+x-3=2+3(x+5) \\\\ 3x^2-8x-3=2+3(x+5) .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 3x^2-8x-3=2+3x+15 \\\\ 3x^2+(-8x-3x)+(-3-2-15)=0 \\\\ 3x^2-11x-20=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 $ 3(-20)=-60 $ and the value of $b$ is $ -11 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-60\}, \{2,-30\}, \{3,-20\}, \{4,-15\}, \{5,-12\}, \{6,-10\}, \\ \{-1,60\}, \{-2,30\}, \{-3,20\}, \{-4,15\}, \{-5,12\}, \{-6,10\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 4,-15 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3x^2+4x-15x-20=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} (3x^2+4x)-(15x+20)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(3x+4)-5(3x+4)=0 .\end{array} Factoring the $GCF= (3x+4) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3x+4)(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} 3x+4=0 \text{ OR } x-5=0 .\end{array} Solving each of the equations above results to \begin{array}{l}\require{cancel} 3x+4=0 \\\\ 3x=-4 \\\\ x=-\dfrac{4}{3} \\\\\text{ OR }\\\\ x-5=0 \\\\ x=5 .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{4}{3},5 \right\} .$
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