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Let $f(x) \in \Bbb R[x]$ and $a \in \Bbb R$. Show that $(x-a)^2 \mid f(x) \iff f(a)=0$ and $f'(a)=0$.

The first implication follows from the product rule. If $(x-a)^2 \mid f(x)$, then $f(x)=(x-a)^2Q(x)$ for some $Q \in \Bbb R[x]$. Now $f(a)=0$ and $f'(x)=2(x-a)Q(x)+(x-a)^2Q'(x)$ and so $f'(a)=0+0=0$.

I don't know how to prove the converse. If I assume that $f(a)=0$ and $f'(a)=0$, then from the first equality I can derive that $(x-a)\mid f(x)$ and from the latter that $(x-a) \mid f'(x)$.

I also can derive that $f(x)=(x-a)P(x)$ for some $P \in \Bbb R[x]$ and that $f'(x)=P(x)+(x-a)P'(x)$, but I don't know how to conclude that the square of $(x-a)$ divides $f$?

Walker
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  • I'd try do it the other way we have $f'(x) = (x-a)Q(x)$ and try to figure out what the antiderivative is. – fleablood Aug 08 '22 at 16:29
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    Hint: change the basis $f(x)=c_0+c_1(x-a)+c_2(x-a)^2+\cdots+c_n(x-a)^n$ then $f(a)=f'(a)=0\implies c_0=c_1=0$ and it is immediate that you can factor $(x-a)^2$. – zwim Aug 08 '22 at 16:41
  • You can consider that $\frac {P(x)}{(x-a)^2} = Q(x) + \frac {R(x)}{(x-a)^2}$ for some $Q(x)$ and $R(x)$ of degree at most $1$. As $(x-a)|P(x)$ we must have $(x-a)|R(x)$ and as $R(x)$ is degree at most $1$, $R(x) =cx+ca$ for some $c\in \mathbb R$. So $P(x) = (x-a)^2Q(x) + cx + ca$. Take the derivative of that. Then if $(x-a)|P'(x)$ it follows that $c= 0$. – fleablood Aug 08 '22 at 16:43
  • .... take zwim's comment which is oodles shorter and clearer than mine.... – fleablood Aug 08 '22 at 16:45

2 Answers2

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Right, so you have: $$f'(x) = P(x)+(x-a)P'(x)$$ Let $x = a$. Then: $$0 = f'(a) =P(a) + 0 =P(a)$$ Since $P$ is a polynomial, it follows that $P(x) = (x-a)Q(x)$. But this means that: $$f(x) = (x-a)^2 Q(x)$$ and we are done.

Abhijeet Vats
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Suppose $f(a) = 0$. Then $(x-a)|f$. Then $f = (x-a)g$ and by the product rule $f' = g + (x-a)g'$. Now if $(x-a)|f'$ this does immediately imply that $(x-a)|g$.

Not much harder to prove: $f$ has a root of order at least $n$ in $a$ exactly if all $f, f', f^{(2)}, \ldots, f^{(n-1)}$ vanish in $a$, with the order is exactly $n$ if $f^{(n)}(a)\neq 0$.

This forms the basis for checking if a polynomial has multiple roots (note that this check can be performed in the field the polynomial is defined on, while the root might live in some algebraic extension of it).

Lazy
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