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.

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

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. First "guess" a zero ${\displaystyle a}${\displaystyle a} of the polynomial ${\displaystyle f}${\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 FML theorem to conclude that ${\displaystyle (x-a)}${\displaystyle (x-a)} is a factor of ${\displaystyle f(x)}${\displaystyle f(x)}.
3. Compute the polynomial ${\displaystyle g(x)=f(x){\big /}(x-a)}${\displaystyle g(x)=f(x){\big /}(x-a)}, for example using polynomial long division.
4. Conclude that any root ${\displaystyle x\neq a}${\displaystyle x\neq a} of ${\displaystyle f(x)=0}${\displaystyle f(x)=0} is a root of ${\displaystyle g(x)=0}${\displaystyle g(x)=0}. Since the polynomial degree of ${\displaystyle g}${\displaystyle g} is one less than that of ${\displaystyle f}${\displaystyle f}, it is "simpler" to find the remaining zeros by studying ${\displaystyle g}${\displaystyle g}.

You wish to find the factors of

${\displaystyle x^{3}+7x^{2}+8x+2.}${\displaystyle 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 ${\displaystyle (x-1)}${\displaystyle (x-1)} is a factor, substitute ${\displaystyle x=1}${\displaystyle x=1} into the polynomial above:

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

${\displaystyle =1+7+8+2}${\displaystyle =1+7+8+2}

${\displaystyle =18.}${\displaystyle =18.}

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

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

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

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

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

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

Let ${\displaystyle f}${\displaystyle f} be a polynomial with complex coefficients, and ${\displaystyle a}${\displaystyle a} be in an integral domain (e.g. ${\displaystyle a\in \mathbb {C} }${\displaystyle a\in \mathbb {C} }). Then ${\displaystyle f(a)=0}${\displaystyle f(a)=0} if and only if ${\displaystyle f(x)}${\displaystyle f(x)} can be written in the form ${\displaystyle f(x)=(x-a)g(x)}${\displaystyle f(x)=(x-a)g(x)} where ${\displaystyle g(x)}${\displaystyle g(x)} is also a polynomial. ${\displaystyle g}${\displaystyle g} is determined uniquely.

This indicates that those ${\displaystyle a}${\displaystyle a} for which ${\displaystyle f(a)=0}${\displaystyle f(a)=0} are precisely the roots of ${\displaystyle f(x)}${\displaystyle f(x)}. Repeated roots can be found by application of the theorem to the quotient ${\displaystyle g}${\displaystyle g}, which may be found by polynomial long division.

• Polynomials
• Mathematical theorems

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