Theoretically, it should work.
Proof for remainder theorem is: $$p(x) = q(x) \cdot (x - a) + r$$ $$p(a) = q(a) \cdot (0) + r$$ $$p(a) = r$$
If the same thing is done for quadratic divisor of ($x^2 - a$): $$p(x) = q(x) \cdot (x^2 - a) + r$$ $$p(\sqrt a) = q(\sqrt a) \cdot (0) + r$$ $$p(\sqrt a) = r$$
Intuitively, I know it wouldn’t because the quotient will have an $x$ term while the evaluation will only get a constant. But, what am I doing wrong?