Usuario:Jgomez53/Esbozos

In algebra, the factor theorem is a theorem for finding out the factors of a polynomial (an expression in which the terms are only added, subtracted or multiplied, e.g. $x^{2}+6x+6$). It is a special case of the polynomial remainder theorem.

The factor theorem states that a polynomial $f(x)$ has a factor $x-k$ if and only if $f(k)=0$.

An example

You wish to find the factors of

$x^{3}+7x^{2}+8x+2$

To do this you would use trial and error finding the first factor. When the result is equal to $0$, we know that we have a factor. Is $(x-1)$ a factor? To find out, substitute $x=1$ into the polynomial above:

$(1^{3})+7(1^{2})+8(1)+2$

This is equal to $18$ not $0$ so $(x-1)$ is not a factor of $x^{3}+7x^{2}+8x+2$. So, we next try $(x+1)$ (substituting $x=-1$ into the polynomial):

$(-1^{3})+7(-1^{2})+8(-1)+2$

This is equal to $0$. Therefore $x-(-1)$, which is to say $x+1$, is a factor, and -1 is a root of $x^{3}+7x^{2}+8x+2$

The next two roots can be found by algebraically dividing $x^{3}+7x^{2}+8x+2$ by $(x+1)$ to get a quadratic, which can be solved directly, by the factor theorem or by the quadratic equation. $(x^{3}+7x^{2}+8x+2) \over (x+1)$ = $x^{2}+6x+2$ and therefore $(x+1)$ and $x^{2}+6x+2$ are the factors of $x^{3}+7x^{2}+8x+2$

Formal version

More formally, it states that for any polynomial $f(x)$, if $a\in R$ satisfies $f(a)=0$, then $f(x)$ can be uniquely written in the form of $f(x)=(x-a)g(x)$ where $g(x)$ is also a polynomial.

This indicates that any $a$ for which $f(a)=0$ is a root of $f(x)$. Double roots can be found by performing polynomial long division.