Descartes' Rule of Signs
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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 Roots 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.
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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 me where the polynomial's zeroes are (I'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 me how many roots I can expect, and of which type.
First, I'll look at the polynomial as it stands, not changing the sign on x. This is the positive-root case:
f (x) = x5 – x4 + 3x3 + 9x2 – x + 5
Ignoring the actual values of the coefficients, I then look at the signs on those coefficients:
f (x) = +x5 – x4 + 3x3 + 9x2 – x + 5
Starting out on this homework, I'll 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. This isn't required, but it'll help me keep track of things while I'm still learning.
Then I count the number of changes:
There are four sign changes in the positive-root case. This number "four" is the maximum possible number of positive zeroes (that is, all the positive x-intercepts) for the polynomial f (x) = x5 – x4 + 3x3 + 9x2 – 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 x-intercepts. 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 positive real zeroes, and there might also be zero (that is, there might be none at all).
I've finished the positive-root case, so now I look at f (–x). That is, having changed the sign on x, I'm now doing the negative-root case:
f (–x) = (–x)5 – (–x)4 + 3(–x)3 + 9(–x)2 – (–x) + 5
= –x5 – x4 – 3x3 + 9x2 + x + 5
I look at the signs:
f (–x) = –x5 – x4 – 3x3 + 9x2 + x + 5
...and I count the number of sign changes:
There is only one sign change in this negative-root 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.
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Using Descartes' Rule of Signs, determine the number of real solutions to:
4x7 + 3x6 + x5 + 2x4 – x3 + 9x2 + x + 1 = 0
I look first at the associated polynomial f (x); using "+x", this is the positive-root case:
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 I look at the polynomial f (–x); using "–x", this is the negative-root case:
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. Then my answer is:
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....
This can be helpful for checking your work. 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.
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Use Descartes' Rule of Signs to find the number of real roots of:
f (x) = x5 + 4x4 – 3x2 + x – 6
First, I look at the positive-root case, which is looking at f (x):
f (x) = +x5 + 4x4 – 3x2 + x – 6
The signs flip three times, so there are three positive roots, or one positive root. Either way, I definitely have at least one positive real root.
Now I look at the negative-root case, which is looking at f (–x):
f (–x) = (–x)5 + 4(–x)4 – 3(–x)2 + (–x) – 6
= –x5 + 4x4 – 3x2 – x – 6
The signs flip twice, so I have two negative roots, or none at all. Then my answer is:
There are three positive roots, or one; there are two negative roots, or none.
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Use Descartes' Rule of Signs to find the number of real roots of:
f (x) = x5 + x4 + 4x3 + 3x2 + x + 1
I look first at f (x):
f (x) = +x5 + x4 + 4x3 + 3x2 + x + 1
There are no sign changes, so there are zero positive roots. Now I 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. Then my answer is:
There are no positive roots, and there are five, three, or one negative roots.
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Use Descartes' Rule of Signs to determine the possible number of solutions to the equation:
x3 + x2 – x – 1 = 0
I'll start with the positive-root case, evaluating the associated functional statement:
f (x) = +x3 + x2 – x – 1
The signs change once, so this has exactly one positive root. Now I'll check the negative-root case:
f (–x) = (–x)3 + (–x)2 – (–x) – 1
= –x3 + x2 + x – 1
The signs switch twice, so there are two negative roots, or else none at all. Then my answer is:
There is exactly one positive root; there are two negative roots, or else there are none.
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Use Descartes' Rule of Signs to determine the possible number of solutions to the equation:
2x4 – x3 + 4x2 – 5x + 3 = 0
I look first at f (x):
f (x) = +2x4 – x3 + 4x2 – 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
= +2x4 + x3 + 4x2 + 5x + 3
There are no sign changes, so there are no negative roots. Then my answer is:
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, suppose the Rational Roots Test gives you a long list of potential zeroes, 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've found every possible negative root (rational or otherwise), so you should now start looking at potential positive roots.
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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