## Intermediate Algebra (12th Edition)

$x=\left\{ -4,7 \right\}$
$\bf{\text{Solution Outline:}}$ 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 $x=\left\{ -4,7 \right\} .$