I think you can still use remainder theorem?
Factorise the divisor into x(x+1)
Using remainder theorem in the original polynomial, we know that:
P(0)=3 and P(-1)=5
Using the polynomial in division form:
P(x)=x(x+1)Q(x) + (ax+3)
We know that
P(-1)=5, so sub it into the divison form to get...