## College Algebra (10th Edition)

$\displaystyle \frac{x-3}{(x+7)}$
$\left[\begin{array}{l|lllll} \text{trinomial} & & & \text{factors of c} & \text{factorization} & \\ ax^{2}+bx+c & b & c & \text{whose sum is b} & & \\ \hline & & & & & \\ x^{2}-3x-10 & -3 & -10 & -5,+2 & (x-5)(x+2) & \\ & & & & & \\ x^{2}+4x-21 & 4 & -21 & +7,-3 & (x+7)(x-3) & \\ & & & & & \\ x^{2}+2x-35 & 2 & -35 & +7,-5 & (x+7)(x-5) & \\ & & & & & \\ x^{2}+9x+14 & 4 & -21 & +7,+2 & (x+7)(x+2) & \end{array}\right]$ ... = $\displaystyle \frac{(x-5)(x+2)(x+7)(x-3)}{(x+7)(x-5)(x+7)(x+2)}$ ... cancel $(x+7),\ (x-5)$ and $(x+2)$ ... = $\displaystyle \frac{x-3}{(x+7)}$