I am a 10th grade student and there is a statement in my math book
If $a$ is a root of the polynomial $f(x)$ then $(x-a)$ is a divisor of $f(x)$
Why is $(x-a)$ a divisor of $f(x)$? Can you please tell me?
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$\begingroup$ Since a is a root it is $f(a)=0$. Therefore your polynomial includes the factor $(x-a)$. You might observe an example. $x^2-2x+1=0$ has the solutions $x_1=-1$ and $x_2=-1$. You can factor it as $(x-1)^2=0$ with the binomic formula. $\endgroup$– MrTopologyJul 6 '16 at 10:44
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$\begingroup$ It's an prompt result of factor theorem $\endgroup$– Zack NiJul 6 '16 at 10:45
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$\begingroup$ Have you learned about synthetic division? If $f(x),g(x)$ are arbitrary polynomials, we can "divide $f$ by $g$" in the sense that we can write $f(x)=g(x)q(x) +r(x)$ where $q(x), r(x)$ are also polynomials and $deg(r(x))<deg(g(x))$. Can you see that this suffices? $\endgroup$– luluJul 6 '16 at 10:46
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$\begingroup$ @TheGreatDuck Why are you saying this??!!! a root of a polynomial is simply a number you plug in that makes it zero, the fact that $(x-\text{root})$ divides the polynomial is a very different story. For instance, one talkes about roots of functions (not necessarily polynomials) where this theorem actually does not hold. $\endgroup$– DanielJul 6 '16 at 23:22
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$\begingroup$ @SolidSnake i forgot that the dividing point is zero. $\endgroup$– user64742Jul 7 '16 at 0:46
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It's a great thing that you feel curiosity for the reasons of the statements that are taught to you!
For this, you have to know a little bit about long division of polynomials. Just like integers, we can divide polynomials, obtaining a quotient and a remainder. More precisely:
Given any polynomials $f$ and $g$, there exist polynomials $q$ (the quotient) and $r$ (remainder) such that $$f = q\cdot g + r$$ and the degree of $r$ is strictly smaller than the degree of $g$.
Now, try to prove your theorem. At first, assume that $a$ is a root of $f(x)$, set $g(x) = x-a$ and apply long division (I'm sure you can do it). The procedure is below, but try to do it by yourself at first.
If we apply long division, you get $q$ and $r$ such that $f = q\cdot (x-a) + r$ and $r$ has degree $0$ (why?), so $r$ is a constant. Since $f(a)=0$, we got $0=f(a)=q(a)\cdot (a-a) + r = 0 + r = r$, so $r=0$ and therefore $f = q\cdot (x-a)$.
The other direction is even easier: if $f(x) = q(x)\cdot(x-a)$, can you see why $f(a)=0$?
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$\begingroup$ This is a verification. But isn't there a proof? I already know about polynomial division and this way of verifying. $\endgroup$ Jul 6 '16 at 17:46
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$\begingroup$ @Adi this is a proof. It's also an 'if and only if' statement or iff. As in $f(a)=0 \text{ iff } (x-a)|f(x)$ $\endgroup$– snultyJul 6 '16 at 19:22
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$\begingroup$ @Adi Why do you say this is a verification? usually, when I think in "verification", I think in particular cases, but we're being very general here. $\endgroup$– DanielJul 6 '16 at 23:19
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$\begingroup$ I guess that he means that the existence of the remainder and quotient is not proved here. $\endgroup$ Jul 7 '16 at 1:34
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Let $$ f (x)=a_n x^n+... +a_1 x+a_0 $$ Suppose $ f (r)=0$. Hence $$ a_n r^n +... + a_1 r +a_0 =0$$
Then $$ f (x)=a_n x^n + ... + a_1 x + a_0 - ( a_n r^n +... + a_1 r +a_0) $$
since the expression between parentheses is zero.
After reordering,
$$ f (x) = a_n (x^n - r^n) + ... + a_1 ( x-r) $$ Note that $$ b^n - t^n= (b-t)(b^{n-1} + b^{n-2} t+... + b t^{n-2}+ t^{n-1})$$ (you can check it?) Hence $$\begin{align} f (x)&= a_n (x-r)(x^{n-1}+...+r^{n-1})+...+a_1 (x-r)\\&= (x-r)(a_n (x^{n-1}+...+r^{n-1})+...+a_1) \end{align}$$ For example, suppose $$ f (x)= a_2 x^2+a_1 x + a_0 $$ and $ f (r)=0$. Hence $$\begin{align} f (x) &= a_2 x^2 + a_1 x + a_0 - ( a_2 r^2 + a_1 r + a_0)\\&= a_2 (x-r)(x+r)+ a_1 (x-r)\\&= (x-r)(a_2 (x+r)+ a_1) \end{align}$$
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2$\begingroup$ I like this answer because it works from first principles and doesn't require the reader to know the factor theorem or remainder theorem or how to do polynomial. (Though Adi, if you're reading, those things are well worth knowing.) $\endgroup$ Jul 6 '16 at 16:47
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This is essentially Factor Theorem, which is a consequence of Remainder Theorem
If you let the polynomial $f(x)$ be represented as $f(x) = (x-a)Q(x) + R$, then you will note that the remainder $R = 0$ if and only if $f(a) = 0$ (i.e. $a$ is a root of $f(x)$). In this circumstance, the polynomial may be represented by $f(x) = (x-a)Q(x)$ and therefore $f(x)$ is divisible by $(x-a)$.
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$\begingroup$ The result is not longer true when $f$ is a polynomial in various variables right? (Working over algebraically closed $k$) for example $y^2 - x^3$ has many roots but can't be factored further. Correct? $\endgroup$– JKEGAug 19 '20 at 0:57
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$\begingroup$ @JKEG The implication by using the notation $f(x)$ is that only univariate polynomials are being considered. This is the usual elementary treatment during an introduction of Factor and Remainder theorems, in fact they generally restrict consideration to real factors only (and often only integer, or at most rational factors). Your question about multivariate polynomials may be better answered here: math.stackexchange.com/questions/1483528/… $\endgroup$– DeepakAug 19 '20 at 4:08
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Recall the factor theorem for polynomials. $$f(x) = (x-a) q(x) + r$$ where $r = f(a)$.
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Lemma $:$ For a field $F, \alpha \in F$ is a root of $a(x)$ if and only if $x-\alpha$ divides $a(x)$
Proof. $(\Longrightarrow)$ Assume that $\alpha$ is a root, i.e., $a(\alpha)=0 .$ Then we can write $a(x)$ as $$ a(x)=(x-\alpha) q(x)+r(x) $$ where $\operatorname{deg}(r(x))<\operatorname{deg}(x-\alpha)=1,$ i.e., $r(x)$ is a constant $r,$ where $$ r=a(x)-(x-\alpha) q(x) $$ Setting $x=\alpha$ in the above equation gives $$ r=a(\alpha)-(\alpha-\alpha) q(\alpha)=0-0 \cdot q(\alpha)=0 $$ Hence $x-\alpha$ divides $a(x)$
($\Longleftarrow$) To prove the other direction, assume that $x-\alpha$ divides $a(x),$ i.e., $a(x)=$ $(x-\alpha) q(x)$ for some $q(x) .$ Then $a(\alpha)=(\alpha-\alpha) q(\alpha)=0,$ i.e., $\alpha$ is a root of $a(x)$
