## Intermediate Algebra (12th Edition)

$x=\left\{ -\dfrac{1}{3},5 \right\}$
$\bf{\text{Solution Outline:}}$ Express the given equation, $(3x+2)(x-3)=7x-1 ,$ 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} 3x(x)+3x(-3)+2(x)+2(-3)=7x-1 \\\\ 3x^2-9x+2x-6=7x-1 \\\\ 3x^2+(-9x+2x-7x)+(-6+1)=0 \\\\ 3x^2-14x-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 $3(-5)=-15$ and the value of $b$ is $-14 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-15\}, \{3,-5\}, \\ \{-1,15\}, \{-3,5\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 1,-15 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3x^2+1x-15x-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} (3x^2+1x)-(15x+5)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(3x+1)-5(3x+1)=0 .\end{array} Factoring the $GCF= (3x+1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (3x+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} 3x+1=0 \text{ OR } x-5=0 .\end{array} Solving each of the equations above results to \begin{array}{l}\require{cancel} 3x+1=0 \\\\ 3x=-1 \\\\ x=-\dfrac{1}{3} \\\\\text{ OR }\\\\ x-5=0 \\\\ x=5 .\end{array} Hence, the solutions are $x=\left\{ -\dfrac{1}{3},5 \right\} .$