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Descartes' Rule of Signs

Descartes' Rule of Signs is a useful help for finding the zeroes of a polynomial, assuming that you don't have the graph to look at. This topic isn't so useful if you have access to a graphing calculator because, rather than having to do guess-n-check to find the zeroes (using the Rational Root Test, Descartes' Rule of Signs, and other tools), you can just look at the picture on the screen.  But if you need to use it, the Rule is actually quite simple.

• Use Descartes' Rule of Signs to determine the number of real zeroes of:
  f (x) = x⁵ – x⁴ + 3x³ + 9x² – x + 5.

Descartes' Rule of Signs will not tell you where the polynomial's zeroes are (you'll need to use the Rational Roots Test and synthetic division, or draw a graph, to actually find the roots), but the Rule will tell you how many roots you can expect. First, look at the polynomial as it stands:

f (x) = x⁵ – x⁴ + 3x³ + 9x² – x + 5

Ignoring the actual values of the coefficients, look at the signs:

f (x) = +x⁵ – x⁴ + 3x³ + 9x² – x + 5

Now note where the signs change from positive to negative or from negative to positive:

Count the number of changes:

There are four sign changes. This number "four" is the maximum possible number of positive zeroes (x-intercepts) for the polynomial  f (x) = x⁵ – x⁴ + 3x³ + 9x² – x + 5. However, remember that some of the roots may be generated by the Quadratic Formula, and these pairs of roots may be complex. Then it may be that certain pairs of roots are not real, and therefore are not graphable as x-intercepts. Because of this possibility, you have to count down by two's. That is, while there may be as many as four real zeroes, there might also be only two, and there might also be zero (none at all).

Now look at f (–x):

f (–x) = (–x)⁵ – (–x)⁴ + 3(–x)³ + 9(–x)² – (–x) + 5

= –x⁵ – x⁴ – 3x³ + 9x² + x + 5

Look at the signs:

f (–x) = –x⁵ – x⁴ – 3x³ + 9x² + x + 5

Count the number of sign changes:

There is only one sign change, so there is exactly one negative root. In this case, you don't count down by two's, because you would end up with a negative number.

There are 4, 2, or 0 positive roots, and exactly 1 negative root.

Some texts have you evaluate  f (x) at x = 1 (for the positive roots) and at x = –1 (for the negative roots), so you would get the expressions "1 – 1 + 3 + 9 – 1 + 5" and "–1 – 1 – 3 + 9 + 1 + 5", respectively. You would analyze the signs in the same manner, however.

The way to keep track of how this Rule works is to note that the number of sign changes for "positive" x (or for when you evaluate at x = +1) give you the number of positive roots. (I say " 'positive' x" in quotes because I don't mean that x itself is positive; I only mean that I haven't put a minus sign on the variable.) The sign changes for "negative" x (or for when you evaluate at x = –1) give you the number of negative roots. There is nothing fancy going on here: you are counting sign changes for positive roots, and then plugging in –x and counting sign changes for negative roots. Here are some more examples:

• Using Descartes' Rule of Signs, determine the number of real solutions to
  4x⁷ + 3x⁶ + x⁵ + 2x⁴ – x³ + 9x² + x + 1 = 0.

Look first at the polynomial  f (x):

f (x) = +4x⁷ + 3x⁶ + x⁵ + 2x⁴ – x³ + 9x² + x + 1

There are two sign changes, so there are two or, counting down in pairs, zero positive solutions. Now look at the polynomial f (–x):

f (–x) = 4(–x)⁷ + 3(–x)⁶ + (–x)⁵ + 2(–x)⁴ – (–x)³ + 9(–x)² + (–x) + 1

= –4x⁷ + 3x⁶ – x⁵ + 2x⁴ + x³ + 9x² – x + 1

There are five sign changes, so there are five or, counting down in pairs, three or one negative solutions.   Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved

There are two or zero positive solutions, and five, three, or one negative solutions.

In the above example, the maximum number of positive solutions (two) and the maximum number of negative solutions (five) added up to the leading degree (seven). This will not always be true, but it will always be true that the sum of the possible numbers of positive and negative solutions will be equal to the degree of the polynomial, or two less, or four less, or.... For instance, if I had come up with a maximum answer of "two" for the possible positive solutions in the above example but had come up with only, say, "four" for the possible negative solutions, then I would known that I had made a mistake somewhere, because 2 + 4 does not equal 7, or 5,  or 3,  or 1.

• Use Descartes' Rule of Signs to find the number of real roots of 
  f (x) = x⁵ + x⁴ + 4x³ + 3x² + x + 1.

Look first at  f (x):

f (x) = +x⁵ + x⁴ + 4x³ + 3x² + x + 1

There are no sign changes, so there are no positive roots. Now look at  f (–x):

f (–x) = (–x)⁵ + (–x)⁴ + 4(–x)³ + 3(–x)² + (–x) + 1

= –x⁵ + x⁴ – 4x³ + 3x² – x + 1

There are five sign changes, so there are as many as five negative roots.

There are no positive roots, and there are five, three, or one negative roots.

• Use Descartes' Rule of Signs to determine the possible number of solutions to the equation 2x⁴ – x³ + 4x² – 5x + 3 = 0.

Look first at f (x):

f (x) = +2x⁴ – x³ + 4x² – 5x + 3

There are four sign changes, so there are four or two or zero positive roots. Now look at  f (–x):

f (–x) = 2(–x)⁴ – (–x)³ + 4(–x)² – 5(–x) + 3

= +2x⁴ + x³ + 4x² + 5x + 3

There are no sign changes, so there are no negative roots.

There are four, two, or zero positive roots, and zero negative roots.

Descartes' Rule of Signs can be useful for helping you figure out (if you don't have a graphing calculator that can show you) where to look for the zeroes of a polynomial. For instance, if the Rational Roots Test gives you a long list of potential zeroes, and you've found one negative zero, and the Rule of Signs says that there is at most one negative root, then you know that you should start looking at positive roots, because there are no more negative roots, rational or otherwise.

Similarly, if you've found, say, two positive solutions, and the Rule of Signs says that you should have, say, five or three or one positive solutions, then you know that, since you've found two, there is at least one more (to take you up to three), and maybe three more (to take you up to five), so you should keep looking for a positive solution.

By the way, in case you're wondering why Descartes' Rule of Signs works, don't. The proof is long and involved; you can study it after you've taken calculus and proof theory and some other, more advanced, classes. I found an interesting paper online (in Adobe Acrobat format) that contains proofs of many aspects of finding polynomial zeroes, and the section on the Rule of Signs goes on for seven pages.

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