Answer
The missin numerator is $2x^2-3x-35$.
Work Step by Step
Factor the numerator and the denominators completely to have:
$\\\dfrac{(x-2)(x+2)}{(x-5)(x-2)} \cdot \dfrac{}{(2x+7)(x+2)}=1$
Cancel the common factors to have:
$\\\require{cancel}\dfrac{\cancel{(x-2)}\cancel{(x+2)}}{(x-5)\cancel{(x-2)}} \cdot \dfrac{}{(2x+7)\cancel{(x+2)}}=1$
For the product to be equal to 1, $x-5$ and $2x+7$ must be canceled out by the factors of the numerator.
Thus, the missing numerator is $(x-5)(2x+7)=2x^2-3x-35$.
