 Solution

Parametrize the line as $\ell (t)=x+t(x^{\prime }-x)$ and set $F(t)=f(\ell (t))$, a polynomial of degree at most three. The Jacobian criterion at a singular point of a hypersurface gives vanishing of its equation and all first derivatives. ConsequentlyThus $t^2$ and $(t-1{)}^2$ both divide $F$. They are coprime in every characteristic, so their degree-four product divides a polynomial of degree at most three. It follows that $F=0$, provingThis is the secant line through singular points of a cubic hypersurface argument. If the given defining polynomial has repeated factors, use its square-free part for the reduced variety's Jacobian criterion: vanishing of that polynomial and its gradient at a singular point implies vanishing of the original polynomial's gradient there as well, by the product rule. The same restriction argument therefore still applies.