Question

What are the possible rational zeros of f(x) = x4 + 2x3 – 3x2 – 4x + 20 ?

Answer 1

Do you have any options?

Answer 2

yes these are the ones from 6.10 they have multiple choice answers can you check that for me please

Answer 3

1, 2, 4, 5, 10, 20

± 1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 7, ± 8, ± 9, ± 10, ± 11, ± 12, ± 13, ± 14, ± 15, ± 16, ± 17, ±18, ± 19, ± 20

± 1, ± 2, ± 4, ± 5, ± 10, ± 20

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20

Answer 4

these are the choices of answers given

Answer 5

trial and error. Descartes' law of signs says that at least two roots are positive, so try +1,+2,+3... until you get an answer. Then use synthetic division to get a lower order polynomial and do it all over again. YAY!

Answer 6

ok thanx

Answer 7

Turing, please explain it man!!

@kiss69, i can't see 6.10

Answer 8

ok so u got it now saifoo legend man

Answer 9

im gonna close it now if not then i will just do it thru open study

Answer 10

get online on facebook for a sec.. plz.

Answer 11

@kiss69, if you are already logged into 6.10, dont close it.. We can't enter in there again.

Answer 12

Well, Descartes', law of signs says that however many sign changes there are in a given polynomial, that's at most how many positive roots you will have. This is x4 + 2x3 – 3x2 – 4x + 20, so the coefficients of sucessive x terms change from + to - twice. Here's a tutorial on that rule
: http://www.purplemath.com/modules/drofsign.htm
once you have a root you can do synthetic division (or polynomial long division) and get a lower order polynomial. Then check its signs and do it again until you get a quadratic.

Answer 13

here is where the sign changes are