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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.