Intermediate Algebra (12th Edition)

$x=\left\{ -4,2 \right\}$
$\bf{\text{Solution Outline:}}$ Express the given equation, $(x-3)(x+5)=-7 ,$ 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(5)-3(x)-3(5)=-7 \\\\ x^2+5x-3x-15=-7 \\\\ x^2+(5x-3x)+(-15+7)=0 \\\\ x^2+2x-8=0 .\end{array} To factor the equation above, find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ Then, express the factored form as $(x+m_1)(x+m_2)=0.$ In the equation above, the value of $c$ is $-8$ and the value of $b$ is $2 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,-8 \}, \{ 2,-4 \}, \\ \{ -1,8 \}, \{ -2,4 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -2,4 \}.$ Hence, the factored form of the equation above is \begin{array}{l}\require{cancel} (x-2)(x+4)=0 .\end{array} Equating each of the factors in the expression above to zero (Zero-Factor Theorem), then, \begin{array}{l}\require{cancel} x-2=0 \text{ OR } x+4=0 .\end{array} Solving each of the equations above results to \begin{array}{l}\require{cancel} x-2=0 \\\\ x=2 \\\\\text{ OR }\\\\ x+4=0 \\\\ x=-4 .\end{array} Hence, the solutions are $x=\left\{ -4,2 \right\} .$