Question

3x^9 –7x^6 + 4x^3 + 15 = 0, state the number of complex roots, the possible number of

real roots, and the possible rational roots.
@phi

Answer 1

@imqwerty

Answer 2

@dan815

Answer 3

@phi

Answer 4

@math&ing001

Answer 5

@pooja195

Answer 6

Read about Descartes' Rule of Signs. Here's a good link : http://www.purplemath.com/modules/drofsign.htm

Answer 7

I know there are 9 passible roots and 2 sign changes but i dont know where to go from there

Answer 8

@pooja195

Answer 9

@eliassaab

Answer 10

@phi

Answer 11

Do you have notes on Descartes rule of signs?

Answer 12

No

Answer 13

@phi

Answer 14

you write down the signs of each term in
+3x^9 –7x^6 + 4x^3 + 15 = 0

in other words, drop everything except the leading sign and you get
+ - + +

now count how many times the sign changes as you move from left to right

Answer 15

2

Answer 16

that means you have 2 positive roots or 0 positive roots
(2 is the max, and you go down by 2 to get the other possibilities)

Answer 17

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These are my choices

Answer 18

if we put in -1 for x, the odd powers of x will be -1 and the even powers are +1

3x^9 –7x^6 + 4x^3 + 15 = 0,
putting in -1 for x we get
-3 -7 -4 +15
or with just the signs
- - - +
there is only 1 sign switch
that is the number of negative roots.

Answer 19

so 2 positive and 1 negative?

Answer 20

the biggest exponent is 9, so there are 9 roots all together
1 is negative (and so real) and up to 2 positive.
Yes I would go with that choice.

Answer 21

so which of the answers would it be?

Answer 22

you have either 1 neg and 0 pos (which means 1 real)
or 1 neg and 2 positive, which means 3 real
there are 9 roots all together.

Answer 23

so c?

Answer 24

if c is the choice that says 1 or 3 real roots.

Answer 25

yes it is

Answer 26

I have a couple more problems that i am not really sure what to do so could you help me with them real quick please?