I posted about an hour ago and i don't know if the answer the guy gave was correct. So im looking for a second opinion.
How do you get the rational zeros? And also how do you find the real zeros? If work could be shown it would be greatly appreciated.
2007-10-11 11:09:01 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I know but if i don't show my work then i don't get credit. I know you can use synthetic division but i don't know what to do with it.
2007-10-11 11:27:52 · update #1
X^6-2x^5-3x^4+4x^3+4x^2
=(x^6-2x^5-3x^4)+(4x^3+4x^2)
=x^4(x^2-2x-3)+4x^2(x+1)
=x^4(x-3)(x+1)+4x^2(x+1)
=(x+1)(x^4(x-3)+4x^2)
2007-10-11 11:18:20 · answer #1 · answered by Anonymous · 1⤊ 0⤋
Obviously X = 0 is a solution since each term has an X in it.
Then factor X^2 out of each term. This leaves you with a quartic equation. These are very hard to factor and to solve.
I would just graph this equation and see where the graph crosses the x axis. This will give you all the real number roots.
It will cross the axis no more than 6 times. One of these places will be at x= 0.
2007-10-11 11:21:27 · answer #2 · answered by Jeffrey K 7 · 0⤊ 0⤋
go on google and paste that in the google browser the answer may pop up
it dont work for some math problems
2007-10-11 11:12:04 · answer #3 · answered by Anonymous · 0⤊ 0⤋
i in my view wish infants might end posting retardedly simplistic homework which would be unable to be defined without purely offering you with the solutions. Ask your mum and dad that might assist you. in the event that they are actually not waiting to discern those out then.... my god...
2016-12-18 05:02:05 · answer #4 · answered by ? 4 · 0⤊ 0⤋
Well I don't know the anwer but do you know the answer?
2007-10-11 11:20:24 · answer #5 · answered by aziza_love_angel 1 · 0⤊ 1⤋