Answer
$x=\left\{ -4,2 \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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\}
.$
