Factor theorem

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.^[1]

The factor theorem states that a polynomial ${\displaystyle f(x)}$ has a factor ${\displaystyle (x-\alpha )}$ if and only if ${\displaystyle f(\alpha )=0}$ (i.e. ${\displaystyle \alpha }$ is a root).^[2]

Factorization of polynomials[]

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:^[3]

1. "Guess" a zero ${\displaystyle a}$ of the polynomial ${\displaystyle f}$. (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 ${\displaystyle (x-a)}$ is a factor of ${\displaystyle f(x)}$.
3. Compute the polynomial ${\textstyle g(x)={\frac {f(x)}{(x-a)}}}$, for example using polynomial long division or synthetic division.
4. Conclude that any root ${\displaystyle x\neq a}$ of ${\displaystyle f(x)=0}$ is a root of ${\displaystyle g(x)=0}$. Since the polynomial degree of ${\displaystyle g}$ is one less than that of ${\displaystyle f}$, it is "simpler" to find the remaining zeros by studying ${\displaystyle g}$.

Example[]

Find the factors of ${\displaystyle x^{3}+7x^{2}+8x+2.}$

Solution: Let ${\displaystyle p(x)}$ be the above polynomial

Constant term = 2

Coefficient of ${\displaystyle x^{3}=1}$

All possible factors of 2 are ${\displaystyle \pm 1}$ and ${\displaystyle \pm 2}$. Substituting ${\displaystyle x=-1}$, we get:

${\displaystyle (-1)^{3}+7(-1)^{2}+8(-1)+2=0}$

So, ${\displaystyle (x-(-1))}$, i.e, ${\displaystyle (x+1)}$ is a factor of ${\displaystyle p(x)}$. On dividing ${\displaystyle p(x)}$ by ${\displaystyle (x+1)}$, we get

Quotient = ${\displaystyle x^{2}+6x+2}$

Hence, ${\displaystyle p(x)=(x^{2}+6x+2)(x+1)}$

Out of these, the quadratic factor can be further factored using the quadratic formula, which gives as roots of the quadratic ${\displaystyle -3\pm {\sqrt {7}}.}$ Thus the three irreducible factors of the original polynomial are ${\displaystyle x+1,}$ ${\displaystyle x-(-3+{\sqrt {7}}),}$ and ${\displaystyle x-(-3-{\sqrt {7}}).}$

References[]