## Intermediate Algebra (12th Edition)

$x=\left\{ 1,6 \right\}$
$\bf{\text{Solution Outline:}}$ Express the given equation, $(x+3)(x-6)=(2x+2)(x-6) ,$ 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} x(x)+x(-6)+3(x)+3(-6)=2x(x)+2x(-6)+2(x)+2(-6) \\\\ x^2-6x+3x-18=2x^2-12x+2x-12 \\\\ (x^2-2x^2)+(-6x+3x+12x-2x)+(-18+12)=0 \\\\ -x^2+7x-6=0 \\\\ -1(-x^2+7x-6)=0(-1) \\\\ x^2-7x+6=0 .\end{array} To factor the trinomial expression above, find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ In the expression above, the value of $c$ is $6$ and the value of $b$ is $-7 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,6 \}, \{ 2,3 \}, \\ \{ -1,-6 \}, \{ -2,-3 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -1,-6 \}.$ Hence, the factored form of the expression above is \begin{array}{l}\require{cancel} (x-1)(x-6)=0 .\end{array} Equating each of the factors in the expression above to zero (Zero-Factor Theorem), then, \begin{array}{l}\require{cancel} x-1=0 \text{ OR } x-6=0 .\end{array} Solving each of the equations above results to \begin{array}{l}\require{cancel} x-1=0 \\\\ x=1 \\\\\text{ OR }\\\\ x-6=0 \\\\ x=6 .\end{array} Hence, the solutions are $x=\left\{ 1,6 \right\} .$