Factor theorem

Short description: A polynomial has a factor (x-k) if and only if k is a root

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 [math]\displaystyle{ f(x) }[/math] has a factor [math]\displaystyle{ (x - k) }[/math] if and only if [math]\displaystyle{ f(k)=0 }[/math] (i.e. [math]\displaystyle{ k }[/math] is a root).

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:

1. "Guess" a zero [math]\displaystyle{ a }[/math] of the polynomial [math]\displaystyle{ f }[/math]. (In general, this can be very hard, but maths 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 [math]\displaystyle{ (x-a) }[/math] is a factor of [math]\displaystyle{ f(x) }[/math].
3. Compute the polynomial [math]\displaystyle{ g(x) = f(x) \big/ (x-a) }[/math], for example using polynomial long division or synthetic division.
4. Conclude that any root [math]\displaystyle{ x \neq a }[/math] of [math]\displaystyle{ f(x)=0 }[/math] is a root of [math]\displaystyle{ g(x)=0 }[/math]. Since the polynomial degree of [math]\displaystyle{ g }[/math] is one less than that of [math]\displaystyle{ f }[/math], it is "simpler" to find the remaining zeros by studying [math]\displaystyle{ g }[/math].

Example

Find the factors of

[math]\displaystyle{ x^3 + 7x^2 + 8x + 2. }[/math]

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 [math]\displaystyle{ (x - 1) }[/math] is a factor, substitute [math]\displaystyle{ x = 1 }[/math] into the polynomial above:

[math]\displaystyle{ x^3 + 7x^2 + 8x + 2 = (1)^3 + 7(1)^2 + 8(1) + 2 }[/math]

[math]\displaystyle{ = 1 + 7 + 8 + 2 }[/math]

[math]\displaystyle{ = 18. }[/math]

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

[math]\displaystyle{ (-1)^3 + 7(-1)^2 + 8(-1) + 2. }[/math]

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

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

[math]\displaystyle{ {x^3 + 7x^2 + 8x + 2 \over x + 1} = x^2 + 6x + 2, }[/math]

and therefore [math]\displaystyle{ (x+1) }[/math] and [math]\displaystyle{ x^2 + 6x + 2 }[/math] are factors of [math]\displaystyle{ x^3 + 7x^2 + 8x + 2. }[/math] Of these the quadratic factor can be further factored using the quadratic formula, which gives as roots of the quadratic [math]\displaystyle{ -3\pm \sqrt{7}. }[/math] Thus the three irreducible factors of the original polynomial are [math]\displaystyle{ x+1, }[/math] [math]\displaystyle{ x-(-3+\sqrt{7}), }[/math] and [math]\displaystyle{ x-(-3-\sqrt{7}). }[/math]