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Recently I've started to study polynomials, when I found out about the remainder and factor theorems as a way to avoid long polynomial division I couldn't understand the reason for every linear factor to start with:

$$x-$$

For example, when the actual x-inteceptor is $-2$, the factor becomes:

$$x+2$$

Why is the $-$ after the $x$?

Maybe its a stupid quastion because if the factor is $x-2$ for example:

$$g(x) = x-2$$

So moving the $-2$ to the left will produce $x=2$(the actual x-interceptor), but I think there is more than that or maybe I just completely misunderstood the all idea.

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  • $\begingroup$ No, it's only that. A polynomial is divisible by $x-\alpha$ if and only $\alpha$ is a root of the polynomial. $\endgroup$
    – Bernard
    Commented Mar 20, 2015 at 18:40
  • $\begingroup$ Thanks @Bernard, I understand, do you think it is more accurate to say that $α$ is the factor for the constant term while $x-α$ is the complete polynomial factor? $\endgroup$
    – Aviel Fedida
    Commented Mar 20, 2015 at 19:46
  • $\begingroup$ No,α can't be considered a factor. $\endgroup$
    – Bernard
    Commented Mar 20, 2015 at 19:54
  • $\begingroup$ @Bernard, I'm not talking about $α$ being the complete polynomial factor, I mean if we have $3x^2-4x+8$ then I can say that $α$ is a factor of the contant term(in our case 8) if the remainder is 0 and $x-α$(again if the remainder is 0) is the complete polynomial factor, am I right? $\endgroup$
    – Aviel Fedida
    Commented Mar 20, 2015 at 20:12
  • $\begingroup$ Being a factor in that sense suppose doing arithmetic (in a rings ofalgebraic integers since the roots are complex). Unfortunately this is not so simple as one mmight hope because we should work in the ring of integers of $\mathbf Q(\mathrm i\sqrt5)$ which is not a factorial ring. $\endgroup$
    – Bernard
    Commented Mar 20, 2015 at 21:34

1 Answer 1

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The factor theorem states that $\,x-a\,$ of a factor of the polynomial $\,f(x)\iff f(a) = 0.$ Negating $\,a\,$ this is instead is $\,x+a\,$ of a factor of the polynomial $\,f(x)\iff f(-a) = 0.$

Notice that when $\,f(x) = g(x)h(x)\,$ then any root of $\,g(x)\,$ is also root of $\,f(x).\,$ Therefore, $ $ when $\,f\,$ has a factor $\,x-a\,$ then its root $\,x = a\,$ is also a root of $\,f,\,$ i.e. $\,f(a) = 0.\,$ And, similarly when $\,f\,$ has a factor $\,x+a,\,$ then its root $\,x = -a\,$ is also a root of $\,f,\,$ i.e. $\,f(-a) = 0$.

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