Question

State

Answer 1

you need to use Descartes Rule of signs.
As its a cubic equation there will be exactly 3 zeros.

there will be either 3 real zeroes OR 1 real and 2 imaginary.

I cannot recall this rule but ill give you a link:

http://hotmath.com/hotmath_help/topics/descartes-rule-of-signs.html

Answer 2

@welshfella

Answer 3

these are my choices: 0 imaginary; 3 real
2 imaginary; 1 real
3 imaginary; 0 real
1 imaginary; 2 real

also, could you expound on how you got that a little more?

Answer 4

well if the some zeroes are complex (imaginary) they will exist as conjugate pairs

e.g. -4i and 4i , or 2+3i and 2 - 3i.

so you cant have 1 or 3 imaginary zeroes.

Answer 5

there are 3 changes of sign in this function (starting with + on the x^3)
+ to -
- to +
+ to -


the link should help you decide if its choice A or B

Answer 6

though its not that good
Try this purple math one:-

http://www.purplemath.com/modules/drofsign.htm

Answer 7

as the last number is 216 - a multiple number - its factors could be roots
try f(3):-

3^3 - 20x^2 + 123x - 216 = 0

so we have 1 root x = 3

Answer 8

So it's A????

Answer 9

@sweetburger

Answer 10

That is correct!

Answer 11

@iwanttogotostanford
After getting 1 real root (x=3), you still have the possibility of 3 real or 1 real and two imaginary.
You can choose between the two after you have done a division by (x-3) to arrive at x^2-17x+72
Check the discriminant of this quadratic. If the roots are real then you have three real roots. If the roots are imaginary (discriminant <0) then you have 1 real and 2 imaginary.