## Intermediate Algebra (12th Edition)

$x=\left\{ 1,4 \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $(x-1)(x+3)=(2x-1)(x-1) ,$ use the FOIL Method and convert the given equation in the form $ax^2+bx+c=0.$ Then express the equation in factored form. Equate each factor to zero using the Zero Product Property. Finally, solve each resulting 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} x(x)+x(3)-1(x)-1(3)=2x(x)+2x(-1)-1(x)-1(-1) \\\\ x^2+3x-x-3=2x^2-2x-x+1 \\\\ (x^2-2x^2)+(3x-x+2x+x)+(-3-1)=0 \\\\ -x^2+5x-4=0 \\\\ -1(-x^2+5x-4)=(0)(-1) \\\\ x^2-5x+4=0 .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $1(4)=4$ and the value of $b$ is $-5 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -1,-4 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x^2-x-4x+4=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^2-x)-(4x-4)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(x-1)-4(x-1)=0 .\end{array} Factoring the $GCF= (x-1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (x-1)(x-4)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions to the equation above are \begin{array}{l}\require{cancel} x-1=0 \\\\\text{OR}\\\\ x-4=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-1=0 \\\\ x=1 \\\\\text{OR}\\\\ x-4=0 \\\\ x=4 .\end{array} Hence, $x=\left\{ 1,4 \right\} .$