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: 33

Answer

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

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ (5x+1)(x+3)=-2(5x+1) ,$ 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} 5x(x)+5x(3)+1(x)+1(3)=-2(5x+1) \\\\ 5x^2+15x+x+3=-2(5x+1) \\\\ 5x^2+16x+3=-2(5x+1) .\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} 5x^2+16x+3=-10x-2 \\\\ 5x^2+(16x+10x)+(3+2)=0 \\\\ 5x^2+26x+5=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 $ 5(5)=25 $ and the value of $b$ is $ 26 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,25\}, \{5,5\}, \\ \{-1,-25\}, \{-5,-5\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 1,25 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 5x^2+1x+25x+5=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} (5x^2+1x)+(25x+5)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(5x+1)+5(5x+1)=0 .\end{array} Factoring the $GCF= (5x+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (5x+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} 5x+1=0 \text{ OR } x+5=0 .\end{array} Solving each of the equations above results to \begin{array}{l}\require{cancel} 5x+1=0 \\\\ 5x=-1 \\\\ x=-\dfrac{1}{5} \\\\\text{ OR }\\\\ x+5=0 \\\\ x=-5 .\end{array} Hence, the solutions are $ x=\left\{ -5,-\dfrac{1}{5} \right\} .$
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