## pinkdollbaby1234 3 years ago 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

1. richyw

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

2. richyw

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

3. pinkdollbaby1234

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

4. richyw

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

5. pinkdollbaby1234

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

6. richyw

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?

7. pinkdollbaby1234

the sign doesnt change anymore cuz all are equal?

8. richyw

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.

9. pinkdollbaby1234

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

10. richyw

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...