Descartes Rule

1. Descartes Rule Descartes' Rule of Signs is a useful help for finding the zeroes of a polynomial

2. 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, synthetic division, and other tools), you can just look at the picture on the screen.

3. 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

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

5. First, I look at the polynomial as it stands, not changing the sign on x, so this is the "positive" case: • Ignoring the actual values of the coefficients, I then look at the signs on those coefficients: f (x) = x5 – x4 + 3x3 + 9x2 – x + 5 f (x) =+x5–x4+ 3x3+ 9x2–x+ 5

6. 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.

7. This number "four" is the maximum possible number of positive zeroes (x-intercepts) for the polynomial f (x) = x5 – x4 + 3x3 + 9x2 – x + 5 there may be as many as four real zeroes, there might also be only two, and there might also be zero (none at all) * I have to count down by two's to find the complete list of the possible number of zeroes.

8. f (–x) =(–x)5 – (–x)4 + 3(–x)3 + 9(–x)2 – (–x) + 5 = –x5 – x4 – 3x3 + 9x2 + x + 5 Now I look at f (–x) (that is, having changed the sign on x, so this is the "negative" case): I look at the signs: f (–x) =–x5–x4– 3x3+ 9x2+x+ 5 ...and I count the number of sign changes:

9. 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.)

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

11. Question: 4x7 + 3x6 + x5 + 2x4 – x3 + 9x2 + x + 1 = 0 Process: the "positive" 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 the "negative" 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.   

12. 4x7 + 3x6 + x5 + 2x4 – x3 + 9x2 + x + 1 = 0 Answer: There are two or zero positive solutions, and five, three, or one negative solutions.

13. 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.