Answer
$x=\left\{ -4,7 \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Express the given equation, $
(2x+1)(x+5)=(x+11)(x+3)
,$ 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}
2x(x)+2x(5)+1(x)+1(5)=x(x)+x(3)+11(x)+11(3)
\\\\
2x^2+10x+x+5=x^2+3x+11x+33
\\\\
2x^2+11x+5=x^2+14x+33
\\\\
(2x^2-x^2)+(11x-14x)+(5-33)=0
\\\\
x^2-3x-28=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 $
-28
$ and the value of $b$ is $
-3
.$
The possible pairs of integers whose product is $c$ are
\begin{array}{l}\require{cancel}
\{ 1,-28 \}, \{ 2,-14 \}, \{ 4,-7 \},
\\
\{ -1,28 \}, \{ -2,14 \}, \{ -4,7 \}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
4,-7
\}.$ Hence, the factored form of the expression above is
\begin{array}{l}\require{cancel}
(x+4)(x-7)=0
.\end{array}
Equating each of the factors in the expression above to zero (Zero-Factor Theorem), then,
\begin{array}{l}\require{cancel}
x+4=0
\text{ OR }
x-7=0
.\end{array}
Solving each of the equations above results to
\begin{array}{l}\require{cancel}
x+4=0
\\\\
x=-4
\\\\\text{ OR }\\\\
x-7=0
\\\\
x=7
.\end{array}
Hence, the solutions are $
x=\left\{ -4,7 \right\}
.$
