- Factor theorem
In

algebra , the**factor theorem**is a theorem for finding out the factors of apolynomial (an expression in which the terms are only added, subtracted or multiplied, e.g. $x^2\; +\; 6x\; +\; 6$). It is aspecial case of thepolynomial 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:: $x^3\; +\; 7x^2\; +\; 8x\; +\; 2\; =\; (1)^3\; +\; 7(1)^2\; +\; 8(1)\; +\; 2$: $=\; 1\; +\; 7\; +\; 8\; +\; 2$: $=\; 18$

As this is equal to 18—not 0—$(x\; -\; 1)$ 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 polynomial with complex coefficients, and $a\; in\; mathbb\{C\}$. Then $f(a)\; =\; 0$

iff $f(x)$ can be written in the form $f(x)=(x-a)g(x)$ where $g(x)$ is also a polynomial. $g$ is determined uniquely.This indicates that those $a$ for which $f(a)\; =\; 0$ are precisely the roots of $f(x)$. Repeated roots can be found by application of the theorem to the quotient $g$, which may be found by

polynomial long division .

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