How to prove that r is a double root if and only if it is a root of a polynomial and of its derivative.

Question

I don't know how to start the question. The title is self explanatory. How to approach and make a formal proof?

Answer 1

Hint $\ $ Wlog, by a shift, assume the root is $\rm\: r = 0.$

Notice $\rm\ \ x^2\: |\ f(x)$

$\rm\ \iff\ \color{#c00}{x\ |\ f(x)}\ $ and $\rm\ x\ \bigg|\ \dfrac{f(x)}{x}$

$\rm\ \iff\ \color{#c00}{f(0)\! =\! 0}\ $ and $\rm\ x\ \bigg|\ \dfrac{f(x)\!-\!\color{#c00}{f(0)}}x\, \left[\!\!\iff\!\! \color{#0a0}{\dfrac{f(x)\!-\!f(0)}x}\bigg|_{\,x\,=\,0} \!= 0\,\right]$

$\rm\ \iff\ f(0)\! =\! 0\ $ and $\rm\ \color{#0a0}{f'(0)}\! =\! 0$

Above we used the Polynomial $\rm\color{#c00}{Factor}$ Theorem and $\rm\color{#0a0}{Derivative}$ Formula.

Remark $\ $ It's often overlooked that many divisibility properties of numbers are specializations of this double-root criterion for (polynomial) functions, e.g. see my answer here., or below from here.

Hint: a conceptual way: we seek $\,\color{#c00}{47^2\mid f(47)}\,$ for $\,f(x) = (x\!+\!5)^{n+1} + ((n\!-\!1)x\!-\!5)(2x\!+\!5)^n$

But $\,f(0)=0=f'(0)\,$ so by the double root test (or first two terms of the Binomial Theorem), it follows that $\,x=0\,$ is a double root so $\,f(x) = x^2 g(x)\,$ for $g$ a polynomial with integer coef's. Evaluating at $\,x=47\,$ yields $\,\color{#c00}{f(47)}= 47^2 g(47).\ $ QED

Answer 2

Hint: (a slightly different approach).

Suppose that $f(x)$ has $r$ as a root $m$ times ($m\geq 1$). Then $f(x)=(x-r)^m g(x)$ for some polynomial $g(x)$ with $g(r)\not=0$. What is $f'(x)$? What has to be true about $m$ if $f'(r)=0$?

Answer 3

I was trying to solve this and came across this question. This is my solution. I hope it helps.

Suppose $P(x)$ is a polynomial of degree n and r is a root of $P(x)$ and its derivative.

$$P(x)=(x-r)Q(x)$$ $$P ^\prime(x)=(x-r)S(x)$$ where $Q(x)$ and $S(x)$ are also polynomials. Based on the first equation, we can also write $$P ^\prime(x)=Q(x)+(x-r)Q^\prime(x)$$ For $x \neq r$ $$S(x)=\frac{P ^\prime(x)}{(x-r)}=\frac{Q(x)+(x-r)Q^\prime(x)}{x-r}$$ $$=\frac{Q(x)}{x-r}+Q^\prime(x)$$ $$=\frac{P(x)}{(x-r)^2}+Q^\prime(x)$$ $$\Rightarrow P(x)=(x-r)^2(S(x)-Q^\prime(x))$$ So, r is a double root of $P(x)$.

If r is a double root of $P(x)$, we have $$P(x)=(x-r)^2T(x)$$ where $T(x)$ is a polynomial. We have $$P^\prime(x)=2(x-r)T(x)+(x-r)^2T^\prime(x)$$ So, r is also a root of the derivative of $P(x)$.