
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 guessncheck to find the zeroes (using the Rational Root Test, Descartes' Rule of Signs, synthetic division, 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.
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, I look at the polynomial as it stands, not changing the sign on x, so this is the "positive" case: f (x) = x^{5} – x^{4} + 3x^{3} + 9x^{2} – x + 5 Ignoring the actual values of the coefficients, I then look at the signs on those coefficients: f (x) = +x^{5} – x^{4}+ 3x^{3}+ 9x^{2}– x+ 5 I draw little lines underneath to highlight where the signs change from positive to negative or from negative to positive from one term to the next: Then I count the number of changes: There are four sign changes in the "positive" case. This number "four" is the maximum possible number of positive zeroes (xintercepts) for the polynomial f (x) = x^{5} – x^{4} + 3x^{3} + 9x^{2} – x + 5. However, some of the roots may be generated by the Quadratic Formula, and these pairs of roots may be complex and thus not graphable as xintercepts. Because of this possibility, I have to count down by two's to find the complete list of the possible number of zeroes. 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 I look at f (–x) (that is, having changed the sign on x, so this is the "negative" case): f (–x) = (–x)^{5} – (–x)^{4} + 3(–x)^{3} + 9(–x)^{2} – (–x) + 5 = –x^{5} – x^{4} – 3x^{3} + 9x^{2} + x + 5 I look at the signs: f (–x) = –x^{5} – x^{4}– 3x^{3}+ 9x^{2}+ x+ 5 ...and I count the number of sign changes: There is only one sign change in this "negative" case, so there is exactly one negative root. (In this case, I don't try to count down by two's, because the first subtraction would give me 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. But you would not simplify, and the numerical values would not be the point; you would analyze only the signs, as shown above.
I look first at the polynomial f (x) (this is the "positive" case): f (x) = +4x^{7} + 3x^{6}+ x^{5}+ 2x^{4}– x^{3}+ 9x^{2}+ x+ 1 There are two sign changes, so there are two or, counting down in pairs, zero positive solutions. Now I look at the polynomial f (–x) (this is the "negative" case): f (–x) = 4(–x)^{7} + 3(–x)^{6} + (–x)^{5} + 2(–x)^{4} – (–x)^{3} + 9(–x)^{2} + (–x) + 1 = –4x^{7} + 3x^{6}– x^{5}+ 2x^{4}+ x^{3}+ 9x^{2}– x+ 1 There are five sign changes, so there are five or, counting down in pairs, three or one negative solutions. Copyright © Elizabeth Stapel 20022011 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). 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 have known that I had made a mistake somewhere, because 2 + 4 does not equal 7, or 5, or 3, or 1.
I look first at f (x): f (x) = +x^{5} + x^{4}+ 4x^{3}+ 3x^{2}+ x+ 1 There are no sign changes, so there are no positive roots. Now I look at f (–x): f (–x) = (–x)^{5} + (–x)^{4} + 4(–x)^{3} + 3(–x)^{2} + (–x) + 1 = –x^{5} + x^{4}– 4x^{3}+ 3x^{2}– 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.
I look first at f (x): f (x) = +2x^{4} – x^{3}+ 4x^{2}– 5x+ 3 There are four sign changes, so there are 4, 2, or 0 positive roots. Now I look at f (–x): f (–x) = 2(–x)^{4} – (–x)^{3} + 4(–x)^{2} – 5(–x) + 3 = +2x^{4} + x^{3}+ 4x^{2}+ 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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