How do you use Descartes' Rule of Signs to describe the possible roots for the function #f(x) = 8x^5-5x^3+x^2-3x+6#?

Question

1371 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-26T08:55:13

Descartes' Rule of Signs gives us that #f(x)# has #4#, #2# or #0# positive real zeros (counting multiplicity) and one negative real zero.

Given:

#f(x) = 8x^5-5x^3+x^2-3x+6#

Note that the pattern of signs of the coefficients is #+ - + - +#. This pattern changes from #+# to #-# or #-# to #+# four times.

By Descartes' Rule of Signs we can deduce that #f(x)# has #4#, #2# or #0# positive real zeros (counting multiplicity).

Then looking at #f(-x)# we find:

#f(-x) = -8x^5+5x^3+x^2+3x+6#

The pattern of signs of coefficients is #- + + + +#. With one change of sign we can deduce that #f(x)# has exactly one negative real zero.

graph{8x^5-5x^3+x^2-3x+6 [-2.5, 2.5, -125, 125]}

In practice, #f(x)# has one negative real zero, no positive real zeros and #4# non-real complex zeros, occuring as two pairs of complex conjugates.