# $f(a) = 0 \implies (x-a) \mid f(x)$? (polynomial zeros imply factors)

Hopefully this is a foolish question with a simple answer.

I've seen it said that, for polynomial $f(x)$

$$f(a)=0 \iff (x-a) \mid f(x)$$

I can see the implication from factor to zero: $(x-a) \mid f(x) \implies f(a)=0$, because anything multiplied by zero is zero, and if $x=a$, then $(x-a) = 0$.

But I'm not sure of the converse: $f(a) = 0 \implies (x-a) \mid f(x)$.

If $f(a)=0$, is the only way it can be zero is if it includes a factor $(x-a)$? How can we be sure there isn't some other circumstance that could make it zero? I guess I'm looking for a proof.

Hopefully it is a foolish question and simple answer!

You can divide $f(x)$ by $(x-a)$ to get $$f(x) = q(x)(x-a) + r(x)$$ where $\deg(r) < \deg(x-a)=1$. Therefore $\deg(r)= 0$ so $r$ is a constant and we can drop the $x$. Now evaluate at $a$: $$f(a) = q(a)(a-a) + r$$

The LHS is zero by assumption, the RHS is equal to $r$. Thus $r=0$ and $(x-a)$ divides $f(x)$ without remainder.

$\,$

Edit: It seems to me like you're confused about why polynomial division always works, so here's the proof to clear things up.

Theorem. For any $f(x), g(x) \in \Bbb Q[x]$, $g(x) \neq 0$, there exist $q(x), r(x) \in \Bbb Q[x]$ such that $$f(x) = q(x)g(x) + r(x)$$ and where $\deg(r) < \deg(g)$.

Proof: First, choose any $q(x), r(x)$ such that the above equation is satisfied; i.e. $q(x) = 0, r(x) = f(x).$ If $\deg(r) < \deg(g)$, we are done.

Therefore assume $\deg(r) \geq \deg(g)$. Let $r(x) = r_nx^n + r_{n-1}x^{n-1} + \dots + r_0$, $g(x) = g_mx^m + g_{m-1}x^{m-1} + \dots + g_0$. Now let $r'(x) = r(x) - x^{n-m}r_ng_m^{-1}g(x)$, let $q'(x) = q(x) + x^{n-m}r_ng_m^{-1}$. Then $\deg r' < \deg r$ (just look at the coefficient of $x_n$) and again we have $$f(x) = q'(x)g(x) + r'(x).$$

If $\deg(r') < \deg(g)$ now, we can stop. If not, repeat the above with $q'$ and $r'$. Because $\deg(r) < \infty$, this procedure must stop after finitely many iterations, so eventually we will have a remainder of degree less than $\deg(g)$.

This concludes the proof.

• Thanks! I can follow this, but don't "get" it. BTW I see that $r(x)$ being a constant is crucial, so that $r(x)=0$ for all x, not just for $x=a$. – hyperpallium May 9 '18 at 8:16
• This proof means that there could be some weird interaction between a polynomial's terms that lead to $f(a)=0$... and if that happens, the terms can be rewritten as having a factor of $(x-a)$. e.g. for $f(x)=x^2-2x$, there's a weird interaction for $x=2$ where the terms cancel $f(2)=2^2-2\cdot 2=0$, which this proof says means $f(x)$ can be written with a factor of $(x-2)$. Which it can be: $x^2-2x = x(x-2)$. This is astonishing to me, partly because it's non-constructive, just showing it must be true, not how. But, using polynomial long division, we can construct the other factor $q(x)$. – hyperpallium May 9 '18 at 8:16
• This proof relies on it always being possible to rewrite $f(x)$ as $q(x)(x-a)+r(x)$, Polynomial Remainder Theorem (PRT). I don't fully "get" polynomial long division, and perhaps getting that will enable me to "get" PRT and the proof given here. I do get the algorithm of polynomial long division, and why it always works - but not why it means $f(x)$ can be written as a term $q(x)b(x)$ with the divisor $b(x)$ as a factor, and a remainder term $r(x)$. It makes sense to multiply the division form by divisor, but not as polynomials (also, divide by zero ($b(x)=(x-a), b(a)$) is undefined). – hyperpallium May 9 '18 at 8:37
• I've added a proof for the existence of a polynomial division algorithm. The resulting $q(x), r(x)$ are unique too. Tell me if this helps. – Lukas Kofler May 9 '18 at 10:24
• It might take less than $\deg(f) - \deg(g) + 1$ steps as we might cancel out not only the highest power but, by chance, the next highest ones as well. It seems like you understood everything else perfectly well though. – Lukas Kofler May 11 '18 at 8:48

Here is an explicit way of showing it. A general polynomial can be written on the form $$f(x) = a_0 + a_1x + a_2 x^2 + \ldots + a_n x^n$$ From this we have that $f(x) - f(a)$ is given by $$f(x) - f(a) = a_1(x-a) + a_2 (x^2-a^2) + \ldots + a_n (x^n - a^n)\tag{1}$$ Now we can try to factor each of the terms above: \begin{align}x^2-a^2 &= (x-a)(x+a)\\x^3-a^3 &= (x-a)(x^2+ax+a^2)\\x^4-a^4 &= (x-a)(x^3+ax^2+a^2x + a^3)\end{align} and in general $x^k - a^k = (x-a)(x^{k-1} + ax^{k-2} + \ldots + a^{k-2}x + a^{k-1})$. This shows that every term on the right hand side of $(1)$ is divisible by $(x-a)$ so in general $(x-a) \mid f(x) - f(a)$. A special case of this is that if $f(a) = 0$ then $(x-a) \mid f(x)$.

• Thanks! In getting the formula $x^k-a^k$, I can guess that when you multiply the series by $x$ and take away the series multiplied by $a$, all the terms in the middle must cancel out, leaving those at the ends - but I can't see how the details actually do that, nor work out how to see it. It seems it should be easy, using $\Sigma$ sums with indices, but I'm not making much progress. – hyperpallium May 9 '18 at 9:10
• @hyperpallium That's exactly how it works. Making a proof for the general case is not that hard: we compute $(x-a)\sum_{k=0}^n x^{n-k}a^{k} = \sum_{k=0}^n x^{n-k+1}a^{k} - \sum_{k=0}^n x^{n-k}a^{k+1}$ and by changing the summation index $k+1 \to K$ in the last sum it becomes $\sum_{K=1}^{n+1} x^{n-K+1}a^{K}$ and you can see that the two sums exactly cancel excepts for $k=0$ in the first term and $K=n+1$ in the last term which leaves us with $x^{n+1} - a^{n+1}$. – Winther May 9 '18 at 9:22
• Thanks! My attempt was similar.... I think I can see it from the difference, $\Sigma - \Sigma$ I'll need to sit down and go through this step by step myself - the complexity is a too much for me, and any little misunderstanding trips me up. The $n$ here is $k-1$ from the answer; the $k$ here is an index. The change of index "$k+1 \to K$" I haven't seen before in concept or notation. – hyperpallium May 12 '18 at 5:47
• Oh! I see, "the last sum" is (of course) the second $\Sigma$, I thought it meant the $\Sigma + -\Sigma$ (since subtraction is an addition of a negated operand... a little knowledge is a dangerous thing). So, the summed amount(?) is identical to the first $\Sigma$, only the indices differ, making it as you say, very clear which ones cancel out. Nice! I end up learning more about the methods than the result - which is what I really need! – hyperpallium May 12 '18 at 5:48