Answer
$x=\left\{ 1,4 \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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\}
.$