## Intermediate Algebra (12th Edition)

$x=\left\{ -4,3 \right\}$
$\bf{\text{Solution Outline:}}$ Express the given equation, $x^2+x-12=0 ,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation. $\bf{\text{Solution Details:}}$ The factored form of the equation above is \begin{array}{l}\require{cancel} (x-3)(x+4)=0 .\end{array} Equating each factor to zero (Zero-Factor Property), then \begin{array}{l}\require{cancel} x-3=0 \text{ OR } x+4=0 .\end{array} Using the properties of equality to solve each of the equation above results to \begin{array}{l}\require{cancel} x-3=0 \\\\ x=3 \\\\\text{ OR }\\\\ x+4=0 \\\\ x=-4 .\end{array} Hence, the solutions are $x=\left\{ -4,3 \right\} .$ $\bf{\text{Supplementary Solution:}}$ To factor the expression, $x^2+x-12 ,$ 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 $(x+m_1)(x+m_2).$ In the given expression, the value of $c$ is $-12$ and the value of $b$ is $1 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,-12 \}, \{ 2,-6 \}, \{ 3,-4 \}, \{ -1,12 \}, \{ -2,6 \}, \{ -3,4 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -3,4 \}.$ Hence, the factored form of the expression above is \begin{array}{l}\require{cancel} (x-3)(x+4) .\end{array}