Associated Topics || Dr. Math Home || Search Dr. Math

### Rational Root Theorem

```
Date: 08/27/98 at 23:40:57
From: Bill Drummond
Subject: Rational zeros

My son doesn't know how to solve the following problem:

List all possible rational zeros of the each function.  Then
determine the rational zeros. A sample problem is:

f(x) = x^3 - 4x^2 + x + 2

We would greatly appreciate your help.

Thanks.
```

```
Date: 08/28/98 at 01:34:57
From: Doctor Pat
Subject: Re: Rational zeros

Bill (and son),

This is called the Rational Root Theorem, and I think it is usually
credited to Descartes. What is says is that if you have a polynomial,
like the one you have above, if it has a rational root (it may not)
then the root must be of a specific type.

To use the rule you need to know that when polynomials are written in
the traditional order, as above with the biggest power first and the
constant last, the number in front of the largest term is called a(n)
and the constant term is called a(0). In your case a(n) = 1 (we usually
don't write a coefficient of 1 in a polynomial) and a(0) is 2.
According to the Rational Root Theorem, the rational root must be a
fraction whose numerator is a factor of a(0) and whose denominator is
a factor of a(n); but they can be positive or negative.

Since the factor of a(n) is only 1, the denominator must be 1 (or -1).
a(0) has two factors -- 1 and 2 so either of these may be the
numerator. The possible choices are then:

2/1, 1/1, -2/1, or -1/1

That is, there are only four possible numbers. The good thing is, if
none of these work, you know for sure there are not any others.

Here is one more example to make sure:

f(x)= 2x^4 + 5x^2 + 7x + 6

Notice that we can ignore all the messy middle stuff and focus on the
2, our a(n) and the 6 which is a(0).

The factors of 2 are 1 and 2, so either of these can be the
denominator. The factors of 6 are 1, 2, 3, and 6 so any of these can be
the numerator.

So what are our choices for possible rational roots:

1/2, 2/2, 3/2, 6/2,
1/1, 2/1, 3/1, 6/1,
or the negatives of any of these

You will probably notice that some of these are duplicates. That is
okay, and we just throw out repeats leaving us with:

1/2, 1, 3/2, 3, 2, 6, and the negatives of each of these.

To check if these are actually roots to our equation, we need to plug
them into the function, and see if we get 0. That's really all there is
to it. Hope that helps you both.

- Doctor Pat, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```

```
Date: 08/28/98 at 17:17:31
From: Doctor Margaret
Subject: Re: Rational zeros

Hi Bill,

Thanks for writing to Dr. Math. It's nice to see a parent taking an
interest in his child's schoolwork. Good for you!

Now on to the task at hand. This assignment comes from a theorem called
the Rational Zeros Theorem. The theorem goes as follows:

Let p be any polynomial function with integer coefficients. The
only rational numbers that can possibly be zeros of p are the
numbers of the form r/s, where r is a divisor of the constant term,
and s is a divisor of the leading coefficient. If none of these
numbers actually turns out to be a zero, then p has no rational
zeros.

Let me explain. A zero of a polynomial (called p(x)) is a number that
when substituted for x will make the polynomial equal to zero. These
numbers were relatively easy to find for polynomials of the form:

p(x) = ax^2 + bx + c

This polynomial, when set to zero becomes a quadratic equation which is
solved by factoring (remember FOIL?) or by using the quadratic formula.

For polynomials of a higher degree (3 and up) there is one method that
is described in the theorem and it works for all polynomials, but
involves some trial and error using synthetic division. We need numbers
of the form r/s, where r is a divisor (factor) of the constant term,
and s a factor of the leading (first) coefficient. In your example, the
constant term is 2, which has factors of +/-1 and +/-2.  The leading
coefficient (of x^3) is 1 and its only factor is +/-1. So now we have:

r    +/-  1,2
--- = ---------
s    +/-   1

Notice how we must also use the negative numbers because the factors
of 1 may be 1 * 1 but may also be -1* -1.  In this case, the possible
zeros are:

+1, -1, +2, -2

Now, I mentioned synthetic division. If we can divide your polynomial
by any of these numbers and come up with no remainder, we have a zero,
the algorithm is as follows to divide a polynomial x of n degrees:

On the top line, at the left, write the alleged zero (call it c for
now). Write all the coefficients of p(x) in order of decreasing powers
of x, include any zero coefficients. Bring down the leading coefficient
to the bottom line.

Multiply each entry on the bottom line by c and add the product to the
next coefficient to get the new entry on the bottom line.

Now watch this. The first n entries on the bottom line are the
coefficients of a new polynomial, call it q(x). The last entry on the
bottom is the remainder. If there is a zero here, then we have a zero
for p(x):

1 | 1  -4  1  2
|______________

1  -3  -2 0

We have a zero so P(1) = 0.

For any zero or root there is a factor. Now we have a kinder, gentler
q(x) = x^2 -3x -2 to work with. This polynomial can now be factored to
get the other zeros. Notice that there will be three zeros for a
polynomial function of the third degree.

Well, that's it. The explanation is a little complicated but the
process is really very simple. If you need extra help with the
synthetic division or if there is something that is not clear, please
write back.

- Doctor Margaret, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Polynomials
Middle School Algebra

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search