Show that a polynomial $P(x)$ has $r$ as a double root if and only if $P'(r)=0$ and $P(r)=0$

Question

Assuming that $r$ is a double root. Then $$P(x)=(x-r)^2\cdot k(x).$$ We also have the derivative: $$P'(x) = 2(x-r)k(x) + (x-r)^2k'(x).$$ Hence, $$P(r) = (r-r^2)k(r)=0$$ and $$P'(r) = 2(r-r)k(r) + (r-r)^2k'(r) = 0.$$ What I cannot show is the converse, that is, assuming that $r$ is a root of $P(x)$ ($P(r)=0$) and $P'(x)$ ($P'(r)=0$); how can we prove that $r$ is a double root of $P(x)$?

Answer 1

Use polynomial division by $(x-r)^2$ to write $$P(x)=(x-r)^2\cdot Q(x)+R(x)$$ where $\deg R<2$, i.e. $R(x)=ax+b$ for some $a,b$. Then $$P'(x)=2(x-r)Q(x)+(x-r)^2Q'(x)+R'(x) $$ From $P(r)=P'(r)=0$, we conclude $R(r)=R'(r)=0$. From $R'(r)=0$ we get $a=0$ and then also $b=0$.

Answer 2

Hint If $r$ is a root of $P(x)$, then $x - r$ is a factor of $P(x)$, that is, $$P(x) = (x - r) Q(x)$$ for some polynomial $Q$. Then, the product rule gives $$P'(x) = (x - r) Q'(x) + Q(x).$$

Now, since $r$ is a root of $P'(x)$, $$0 = P'(r) = (r - r) Q'(x) + Q(r) = Q(r),$$ that is, $r$ is also root of $Q$. So, $$Q(x) = (x - r) R(x)$$ for some polynomial $R(x)$, and thus $$P(x) = (x - r) Q(x) = (x - r) [(x - r) R(x)] = (x - r)^2 R(x). \,\,\square$$ The same approach works for roots of arbitrary order.

Answer 3

Let $Q(x) = P(x+r)$. Then $Q$ has a double root at zero exactly when the constant and linear terms vanish, i.e. when $Q(0) = Q'(0) = 0$. The result follows from the chain rule.