i dont understand how to do this

use descartes rule of signs to describe the root of
h(x)=4x^4-5x^3+2x^2-x+5

Question

i dont understand how to do this

use descartes rule of signs to describe the root of
h(x)=4x^4-5x^3+2x^2-x+5

richyw · Answer

$$h(x)=4x^4-5x^3+2x^2-x+5$$how many times does it change signs?

richyw · Answer

i see 4 now do it for $h(-x)$...

anonymous · Answer

okay so what do you mean by that  solve for h?

richyw · Answer

no $$h(-x)=4(-x)^4-5(-x)^3+2(-x)^2-(-x)+5$$Now simplify that, and then count the sign changes.

anonymous · Answer

oh okay so i would plug in 4 for -x  since it changes four times or is im wrong again?

richyw · Answer

yeah you are wrong. this is what you should get.$$h(-x)=4x^4+5x^3+2x^2+x+5$$Now how many times does the sign change places there?

anonymous · Answer

the sign doesnt change anymore cuz all are equal?

richyw · Answer

yup so you can conclude that there are no negative roots, and either 4, 2 or 0 positive roots. You also know that there are 4 roots total. so you can either have 4 positive real and 0 complex,  2 and 2, or 0 real positive roots and 4 complex roots.

anonymous · Answer

omg thay seem soo simple i have to practice on this more but thankyou so much u help me understand it a bit

richyw · Answer

yup no worries! And in this case that is all it can tell you. Sometimes it can tell you exactly how many of each there are, but sometimes it gives you a few options. It is still useful though because you can use it before you use other methods to get an idea what you are looking for...