Answer to Question #186426 in Algebra for Jocelyn

Question

h(x)=3x^8-5x^4-x+2 descartes rule of signs

Answer 1

$h(x)=3x^8-5x^4-x+2$

using h(+x)

$h(x)=+3x^8-5x^4-x+2$

there are 2 sign changes

This means there are two or no positive roots.

and using h(-x)

$h(x)=3(-x)^8-5(-x)^4-(-x)+2$

$=+3x^8-5x^4+x+2$

there are 2 sign changes

This means that there are two or no negative roots.

There are two or zero positive roots, and two or zero negative roots.