Thursday, October 11, 2018

College Algebra, Chapter 4, 4.4, Section 4.4, Problem 68

Determine how many positive and how many negative real zeros the polynomial P(x)=x8x5+x4x3+x2x+1 can have, using the Descartes' Rule of signs. Then determine the possible total number of real zeros.

We first look at P(x)

P(x)=+x8x5+x4x3+x2x+1

P(x) has six variations in sign, so it has either 6,4,2 or positive roots.

Now,


P(x)=(x)8(x)5+(x)4(x)3+(x)2(x)+1P(x)=+x8+x5+x4+x3+x2+x+1


So P(x) has no variation in sign. Thus, P(x) has negative roots, making a total of either 7,5,3 or 1 real zeros. Since is a zero but is neither positive nor negative.

No comments:

Post a Comment