Nernst effect

In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. 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 $f(k)=0$ (i.e. $k$ is a root).

Factorization of polynomials

Mining Engineer (Excluding Oil ) Truman from Alma, loves to spend time knotting, largest property developers in singapore developers in singapore and stamp collecting. Recently had a family visit to Urnes Stave Church. Two problems where the factor theorem is commonly applied are those of factoring a polynomial and finding the roots of a polynomial equation; it is a direct consequence of the theorem that these problems are essentially equivalent.

The factor theorem is also used to remove known zeros from a polynomial while leaving all unknown zeros intact, thus producing a lower degree polynomial whose zeros may be easier to find. Abstractly, the method is as follows:

1. "Guess" a zero $a$ of the polynomial $f$. (In general, this can be very hard, but math textbook problems that involve solving a polynomial equation are often designed so that some roots are easy to discover.)
2. Use the factor theorem to conclude that $(x-a)$ is a factor of $f(x)$.
3. Compute the polynomial $g(x)=f(x){\big /}(x-a)$, for example using polynomial long division or synthetic division.
4. Conclude that any root $x\neq a$ of $f(x)=0$ is a root of $g(x)=0$. Since the polynomial degree of $g$ is one less than that of $f$, it is "simpler" to find the remaining zeros by studying $g$.

Example

Find the factors at

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

To do this you would use trial and error to find the first x value that causes the expression to equal zero. To find out if $(x-1)$ is a factor, substitute $x=1$ into the polynomial above:

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

$=1+7+8+2$

$=18.$

As this is equal to 18 and not 0 this means $(x-1)$ 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

Let $f$ be a one-variable polynomial with coefficients in a commutative ring $R$, and let $a\in R$. Then $f(a)=0$ if and only if $f(x)=(x-a)g(x)$ for some polynomial $g$. In this case, $g$ is determined uniquely.

As for the problem of algorithmically finding all roots, if $f$ is given and $a$ is known, then $g$ can be computed by polynomial long division; then one can compute the remaining roots of $f$, including repeated roots, by factoring $g$.

References

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