Constructing Jacobians of rank 1

Let be a number field, let g>1 be an integer and let f(x) be a polynomial that splits into 2g+1 distinct linear factors. Write C for the hyperelliptic curve given by C:y^2=f(x) and write J=Jac(C) for its Jacobian. Under mild technical assumptions on f that are satisfied almost always, we prove that there exists some d in K^{\times} such that the quadratic twist J_d has rank exactly equal to 1. As a consequence, we deduce that for any positive integer g, there exists an absolutely simple abelian variety over K with dimension equal to g and rank equal to 1.