Question

find all real solutions of x^3−2x^2−3x+6=0

Answer 1

x=2 is a solution so (x-2) is a factor. Divide synthetically to get a quadratic expression, which can be solved by either factoring or the quadratic formula.

Answer 2

let f(x)=x^3−2x^2−3x+6
since f(2)=0
x-2 must be factor of f(x)
use synthetic divsion or simple division you'll get than quadrtic equation. then solve that eqatin

Answer 3

what told me that f(x) is 2

Answer 4

nothing really. Trial and error. Descartes' Rule of signs says there are at most 2-positive roots, so I just plugged in x=1 and x=2 and got lucky.

Answer 5

x=2 and\[\pm \sqrt{3}\]

Answer 6

it's trial method, just put such value of x at which f(x)-0