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