3.3. Apply Descartes\'s rule of signs to count the possible number of positive,

Question

3.3. Apply Descartes's rule of signs to count the possible number of positive, negative, and complex roots (organize a table) 2 p(x)=x-5x +5x +5x-6 Select one sible| Possible | Possible | Total Positive Negative Complex |Number RootsRoots Roots of Roots 4 Possible | Possible Possible Total Positive Negative Complex Number Roots Roots Roots of Roots 4 Possible Possible | Possible|Total Positive Negative Complex Number RootsRoots Roots of Roots 4 4 sible| Possible | Possible | Total Positive Negative Complex Number Roots Roots Roots of Roots 4

Explanation / Answer

Descartes' Rule of Signs will not tell us where the polynomial's zeroes are (we'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 us how many roots I can expect, and of which type.

First, we'll look at the polynomial as it stands, not changing the sign on x. This is the positive-root case:

F(x) = x4-5x3+5x2+5x-6

Ignoring the actual values of the coefficients, we then look at the signs on those coefficients:

F(x) = x4-5x3+5x2+5x-6

See 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 us keep track of things while we are still learning.

F(x) = +x4-5x3+5x2+5x-6

Then we count the number of changes:

F(x) = +x4-5x3+5x2+5x-6

There are three sign changes in the positive-root case. This number "three" is the maximum possible number of positive zeroes (that is, all the positive x-intercepts) for the polynomial

F(x) = +x4-5x3+5x2+5x-6

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.  

We've finished the positive-root case, so now I look at F (–x). That is, having changed the sign on x, we are now doing the negative-root case:

F(- x) = (-x4) - 5(- x3)+5 (- x2) + (- 5x) - 6

= x4 + 5x3 + 5x2 - 5x - 6

I look at the signs and I count the number of sign changes:

F(- x) = x4 + 5x3 + 5x2 - 5x - 6

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 3 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 - 5 + 5 + 5 - 6" and "1 + 5 +5 - 5 - 6", respectively. But you would not simplify, and the numerical values would not be the point; you would analyze only the signs.

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