Given $a \in F$ and $f \in F[x]$, show that $(X-a)^2$ is a divisor of $f$ iff $a$ is a root of $f$ and a root of the derivative of $f$.

Here's what I don't understand -- How to prove in regards to the derivative?

This was my proof so far:

Assume $(X-a)^2$ is a divisor but $a$ is not a root, therefore:

$f = q*(X-a)^2 + r$ where $r = 0$. Also $f(a) \neq 0$

Now, if we test for $a$:

$f(a) = q(a)(a-a)^2 + 0 = q(a)*0^2 \neq 0$

$0 \neq 0$

That's a contradiction, therefore a has to be $a$ root of $f$

  • $\begingroup$ Relevant: math.stackexchange.com/questions/976749/… $\endgroup$ – Richard D. James Mar 29 at 7:12
  • $\begingroup$ You don't really need the proof by contradiction here. You can just start with $(X-a)^2$ is a divisor of $f$ and conclude that $f(a)=0$ with nearly exactly the same steps. $\endgroup$ – aschepler Mar 29 at 15:26

Let $(X-a)^2$ is a divisor of $f$, then $f=(X-a)^2q$ for some polynomial $q$. Now of course $f(a)=0$. Also $$f'(x)=2(X-a)q+(X-a)^2q'\implies f'(a)=0.$$

Now for the converse, suppose $a$ is a root of $f$ and a root of $f'$. From $f(a)=0$ we have $f=(X-a)q$ and therefore $$f'=q+(X-a)q'$$ From $f'(a)=0$ we see that $$0=f'(a)=q(a)$$ thus $q=(X-a)p$ for some polynomial $p$, therefore $$f=(X-a)q=(X-a)^2p$$

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  • $\begingroup$ Could you briefly explain line 2? Why does $f=(X-a)q$? $\endgroup$ – Alex Osheter Mar 29 at 7:15
  • $\begingroup$ $f$ being a polynomial with root $a$, means that after dividing $f$ by $X-a$ we have remainder $=0$, i.e. $f=(X-a)q+r$ with $r=0$ $\endgroup$ – Qurultay Mar 29 at 7:17
  • $\begingroup$ Oh, right. Thank you! But I didn't entirely understand what you did. You started with the assumption that $(X-a)^2$ is a divisor of $f$ and arrived at the same conclusion in the end... If we prove that $a$ has to be a root of $f$ then, wouldn't your line 1 proof be enough? $f'(a)=0$. If we assume $f'(a) \neq 0$ we'll get a contradiction, and done - no? $\endgroup$ – Alex Osheter Mar 29 at 7:23
  • $\begingroup$ From $f'(a)\ne0$, how we conclude $f(a)\ne 0$? For example for $f=x^2-1$, we have $f'(1)=2\ne0$ but $f(1)=0$ $\endgroup$ – Qurultay Mar 29 at 7:32
  • $\begingroup$ Oh okay, thank you. I understand! $\endgroup$ – Alex Osheter Mar 29 at 7:35

Suppose that

$(X - a)^2 \mid f(X) \in F[X]; \tag 1$


$\exists g(X) \in F[X], \; f(X) = (X - a)^2 g(X); \tag 2$

we have

$f(a) = (a - a)^2 g(a) = 0g(a) = 0, \tag 3$


$f'(X) = 2(X - a)g(X) + (X - a)^2 g'(X), \tag 4$


$f'(a) = 2(a - a)g(a) + (a - a)^2g'(a) = 2 \cdot 0 \cdot g(a) + 0 \cdot g'(a) = 0; \tag 5$

thus $a$ is a root of both $f(X)$ and $f'(X)$.

Conversely, if

$f'(a) = f(a) = 0, \tag 6$


$\exists g(X) \in F[X], \; f(X) = (X - a)g(X), \tag 7$


$f'(X) = g(X) + (X - a)g'(X); \tag 8$


$g(a) = g(a) + (a - a)g'(a) = f'(a) = 0; \tag 9$

this in turn implies

$\exists h(X) \in F[X], \; g(X) = (X - a)h(X), \tag{10}$

so that

$f(X) = (X - a)g(X) = (X - a)^2h(X), \tag{11}$

and we conclude that

$(X - a)^2 \mid f(X). \tag{12}$

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