## College Algebra (10th Edition)

$\displaystyle \frac{(x-2)(x+3)(x-5)}{(x+5)(x-1)(x-3)}$
$\left[\begin{array}{llllll} \text{trinomial} & & & \text{factors of c} & \text{factorization} & \\ ax^{2}+bx+c & b & c & \text{whose sum is b} & & \\ \hline & & & & & \\ x^{2}+x-6 & 1 & -6 & -2,+3 & (x-2)(x+3) & \\ & & & & & \\ x^{2}-25 & 0 & -25 & +5,-5 & (x+5)(x-5) & \\ & & & & & \\ x^{2}+4x-5 & 4 & -5 & +5,-1 & (x+5)(x-1) & \\ & & & & & \\ x^{2}+2x-15 & 2 & -15 & -3,+5 & (x-3)(x+5) & \end{array}\right]$ ... = $\displaystyle \frac{(x-2)(x+3)(x+5)(x-5)}{(x+5)(x-1)(x-3)(x+5)}$ ... cancel one $(x+5)$ from both sides of the fraction line ... = $\displaystyle \frac{(x-2)(x+3)(x-5)}{(x+5)(x-1)(x-3)}$