Rational Roots Theorem

March 12, 2010

This is a discussion on the method of the Rational Roots Theorem. This method allows you to find all of the rational roots of any polynomial with rational (integer) coefficients.

To start, you must set up your polynomial appropriately. If your coefficients are irrational or complex, you must factor our any common coefficient, in particular the irrational and complex components, in order to make all coefficients rational and real. Multiplying the polynomial on whole by an arbitrary irrational/complex constant is ultimately irrelevant. But in order to apply the theorem, all coefficients of the polynomial you are finding the roots to must be rational and real. This theorem cannot be applied if your coefficients cannot be rationalized.

In fact, the coefficients must all be integers in order to apply the theorem. So if your coefficients are non-integers but rational then multiply through by a the common denominator. Multiplying the polynomial by a constant in no way affects the roots themselves, but it may be necessary to make all coefficients integers.

We should ultimately end up with a polynomial of the form:

$C\left(a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \right)$

Where all $a_k$ are integers and $C$ is an arbitrary constant (irrational or complex is fine).

Notice that the roots of this polynomial occur when it is set equal to zero.

$C\left(a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \right) = 0$

Thus the constant C drops out and is irrelevant.

$a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0$

The theorem also assumes that there is a constant term. That is to say, $a_0 \ne 0$ . If it is equal to zero then $x=0$ is a root and $x$ is a factor. It should be noted and dropped out to produce a new polynomial with a decremented degree having a constant term.

The theorem is simple from here on out. All you need to focus on is the lead coefficient $a_n$ and the constant term $a_0$

Applying the Theorem

You must be able to find the factors of a constant. Not the prime factors exclusively, but all factors. All integers, composite or prime, that can divide into a constant.

Create a set of numbers which form the integer factors of your lead coefficient. Assume for a moment that $a_n = 4$ , then the list of integer factors are $\{1,2,4\}$ .

Create a second set of numbers which form the integer factors of the constant term. Assuming $a_0 = 6$ then the set of integer factors is $\{1, 2, 3, 6\}$ .

The next step is to take all rational numbers that can be constructed from each possible combination, in which the second set forms the numerator and the first set forms the denominator.

That is, each number from the set constructed of the factors of the constant term divided by each number from the set constructed of the factors of the lead coefficient.

Taking each combination:

$\{\frac{1}{1},\frac{2}{1},\frac{3}{1},\frac{6}{1},\frac{1}{2},\frac{2}{2},\frac{3}{2},\frac{6}{2},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{6}{4}\}$

This sets components can be reduced and the set consolidated.

$\{\frac{1}{4},\frac{1}{2},\frac{3}{4},1,\frac{3}{2},2,3,6\}$

This is pretty simple so far, eh?

Now we take the positive and negative of each of those values:

$\{\pm\frac{1}{4},\pm\frac{1}{2},\pm\frac{3}{4},\pm1,\pm\frac{3}{2},\pm2,\pm3,\pm6\}$

Or simply:

$\{-6,-3,-2,-\frac{3}{2},-1,-\frac{3}{4},-\frac{1}{2},-\frac{1}{4},\frac{1}{4},\frac{1}{2},\frac{3}{4},1,\frac{3}{2},2,3,6\}$

This set contains all of the possible rational roots that the polynomial can have. The polynomial cannot have rational roots outside of this set.

Any generic roots that the polynomial may have that is not accounted for by the Rational Roots Theorem fall in the realm of non-rational numbers. Roots that exist outside of the set produced by the theorem are either complex or real-irrational.

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Posted in algebra, method, polynomial | Tagged algebra, method, polynomial, roots, theorem, zeros |

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[...] restrict the real domain significantly. It can eliminate possible rational roots derived from the Rational Roots Theorem so that they dont need to be tested. It can narrow the real domain far more restrictively than [...]
by Upper and Lower Bounds Test « Cogito's Math Tutorage March 13, 2010 at 5:04 pm

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