## Intermediate Algebra (12th Edition)

$x=\left\{ -\dfrac{1}{2},6 \right\}$
$\bf{\text{Solution Outline:}}$ 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 $x=\left\{ -\dfrac{1}{2},6 \right\} .$