Question

Please help.
Use Descartes' rule of sign to describe the roots of h(x)=4x^4-5x^3+2x^2-x+5

Answer 1

@Loser66

Answer 2

Hope it help you

Answer 3

lol.
you refers to ASKER

Answer 4

do you still need help?

Answer 5

Can you help me @triciaal

Answer 6

need help

Answer 7

go on this http://www.mathwords.com/d/descartes_rule_of_signs.htm

Answer 8

These are
+4, -5, +2, - 1, + 5. Next focus on the sign
changes as you go along from left to right.
We are going to count them. Bracket them as follows:
(+4, -5)→(-5, +2)→(+2, -1)→(-1, + 5). So as you go
along the sign has changed 4 times. Descartes' rule
of signs tells us that the number of positive roots is
either 4 or 4 less an even number, I.e., 4 0r 4 - 2 = 2
or 4 - 4 = 0. So there are 0, 2 or 4 +ve roots. IT DOES NOT
TELL US WHICH OF 0, 2, 4 IT IS. Now for the -ve roots.
These are just the number of +ve roots of h(-x), which is
4x⁴+ 5x³ + 2x² +x +5. There are no changes of sign here,
so h(-x) has 0 +ve roots, i.e., h(x) has 0 -ve roots. Thus, if
none of the roots are complex all its 4 roots must be +ve,
but we don't know this. So all we can know from the rule
of signs is that the polynomial has no -ve roots and 0, 2 or
4 +ve ones.

Answer 9

is she there or nah

Answer 10

I'm here

Answer 11

Thank you guys so much!

Answer 12

4x4−5x3+2x2−x=−5

Answer 13

yw

Answer 14

anytime

Answer 15

lol