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