Answer
$x=\left\{ -\dfrac{1}{2},6 \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
2(x-1)^2-7(x-1)-15=0
,$ use substitution and then the method on factoring trinomials. Then equate each factor to zero (Zero Product Property) and solve for the value of the variable. Finally use back-substitution and then solve for the value of the original variable.
$\bf{\text{Solution Details:}}$
Let $z=x-1.$ Then the equation above is equivalent to
\begin{array}{l}\require{cancel}
2z^2-7z-15=0
.\end{array}
To factor the trinomial expression above, note that the value of $ac$ is $
2(-15)=-30
$ and the value of $b$ is $
-7
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-30\}, \{2,-15\}, \{3,-10\}, \{5,-6\},
\\
\{-1,30\}, \{-2,15\}, \{-3,10\}, \{-5,6\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
3,-10
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
2z^2+3z-10z-15=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}
(2z^2+3z)-(10z+15)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
z(2z+3)-5(2z+3)=0
.\end{array}
Factoring the $GCF=
(2z+3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(2z+3)(z-5)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then the solutions are
\begin{array}{l}\require{cancel}
2z+3=0
\\\\\text{OR}\\\\
z-5=0
.\end{array}
Since $z=x-1,$ by back-substitution, the solutions are
\begin{array}{l}\require{cancel}
2(x-1)+3=0
\\\\\text{OR}\\\\
(x-1)-5=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
2(x-1)+3=0
\\\\
2x-2+3=0
\\\\
2x+1=0
\\\\
2x=-1
\\\\
x=-\dfrac{1}{2}
\\\\\text{OR}\\\\
(x-1)-5=0
\\\\
x-1-5=0
\\\\
x-6=0
\\\\
x=6
.\end{array}
Hence, the solutions are $
x=\left\{ -\dfrac{1}{2},6 \right\}
.$
