# Help with proof of polynomial divisor

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 $$f$$

• – Richard D. James Mar 29 at 7:12
• 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. – 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 $$f=(X-a)q=(X-a)^2p$$

• Could you briefly explain line 2? Why does $f=(X-a)q$? – Alex Osheter Mar 29 at 7:15
• $f$ 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 $r=0$ – Qurultay Mar 29 at 7:17
• 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? – Alex Osheter Mar 29 at 7:23
• 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 $f(1)=0$ – Qurultay Mar 29 at 7:32
• Oh okay, thank you. I understand! – Alex Osheter Mar 29 at 7:35

Suppose that

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

then

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

we have

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

and

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

whence

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

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

Conversely, if

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

then

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

whence

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

then

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

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

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