Tangent line at $x_1$ to polynomial curve $p(x)$ of degree at least $2$ implies $x_1$ is a double root of $p(x) - p^{'}(x_1)(x-x_1)$?

Question

Tangent line at $x_1$ to polynomial curve $p(x)$ of degree at least $2$ implies $x_1$ is a double root of $p(x) - p^{'}(x_1)(x-x_1)$ ?.

Suppose I have a polynomial function $p(x): \mathbb R \rightarrow \mathbb R$ given by $x \mapsto x^3 + Ax + B$.

Also suppose $L:p^{'}(x_1)(x-x_1)$ is tangent line at the point $x_1$.

Is it always true that $x^3 + Ax + B - p^{'}(x_1)(x-x_1)$ has a double root $x_1$ ?

Is it possible to make this more general ?

Answer 1

Yes. Let $f$ be differentiable and let $g(x)=mx+b$ be the equation for the tangent line at $x_0$, that is $g(x_0)=f(x_0)$ and $m = g'(x_0) = f'(x_0)$. Then we consider the function $h(x)=f(x)-g(x)$ and observe that $h(x_0)=f(x_0)-g(x_0)=0$ and $h'(x_0)=f'(x_0)-g'(x_0)=0$ by the very conditions given. Recall that a polynomial (if $h$ is one) has a (at least) $k$fold root at $x_0$ if $h(x_0)=h'(x_0)=\ldots =h^{(k-1)}(x_0)=0$.