= Solution
Parametrize the line as $\ell(t)=x+t(x'-x)$ 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. Consequently
$$
F(0)=F(1)=0,\qquad
F'(0)=\sum_j(x'_j-x_j)f_{t_j}(x)=0,\qquad
F'(1)=\sum_j(x'_j-x_j)f_{t_j}(x')=0.
$$
Thus $t^2$ 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 $F=0$, proving
$$
\boxed{\ell(\mathbb A^1)\subseteq X.}
$$
This 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.
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