# 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 ${\displaystyle (x-k)}$ if and only if ${\displaystyle f(k)=0}$ (i.e. ${\displaystyle k}$ 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

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

To do this one would use trial and error (or the rational root theorem) to find the first x value that causes the expression to equal zero. To find out if ${\displaystyle (x-1)}$ is a factor, substitute ${\displaystyle x=1}$ into the polynomial above:

{\displaystyle {\begin{aligned}x^{3}+7x^{2}+8x+2&=(1)^{3}+7(1)^{2}+8(1)+2\\&=1+7+8+2\\&=18\end{aligned}}}

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

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

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

The next two roots can be found by algebraically dividing ${\displaystyle x^{3}+7x^{2}+8x+2}$ by ${\displaystyle (x+1)}$ to get a quadratic:

${\displaystyle {x^{3}+7x^{2}+8x+2 \over x+1}=x^{2}+6x+2,}$

and therefore ${\displaystyle (x+1)}$ and ${\displaystyle x^{2}+6x+2}$ are factors of ${\displaystyle x^{3}+7x^{2}+8x+2.}$ 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

1. ^ Sullivan, Michael (1996), Algebra and Trigonometry, Prentice Hall, p. 381, ISBN 0-13-370149-2.
2. ^ Sehgal, V K; Gupta, Sonal, Longman ICSE Mathematics Class 10, Dorling Kindersley (India), p. 119, ISBN 978-81-317-2816-1.
3. ^ Bansal, R. K., Comprehensive Mathematics IX, Laxmi Publications, p. 142, ISBN 81-7008-629-9.