Question

Try for fun.
Solve for x,
Given that x^4-2x^3-5x^2+10x-3=0.

Answer 1

It gonna have 4 solutions..

Answer 2

ok, then,

Answer 3

the easiest way is to factorize it .. into lower terms. best of luck @Eyad

Answer 4

if you factorize it to the lowest power of x^2 you'll get x=3 or -7 or (2+\[\sqrt{24})/2\]or (2-\[\sqrt{24})/2\]

Answer 5

\[x^4-2x^3-5x^2+10x-3=0.\]

Answer 6

now looking good:)

Answer 7

thank u @maheshmeghwal9

Answer 8

it's really a fun:)

Answer 9

yw:)

Answer 10

grr, no rational roots..
Descartes' rule shows 3 positive roots and 1 negative root.
From plotting a few points, there are roots -3<x<-2, 0<x<1, 1<x<2, and 2<x<3.
Going to have to bring out the big guns to find those irrationals. ;-)