what are the rational roots of x^3-2x^2+x+18=0

Answer 1

According to the rational roots theorem, any rational roots of this polynomial are expressible in the form p/q for integers 

p,q with p a divisor of the constant term 18 and q a divisor of the coefficient 1 of the leading term.

That means that the only possible rational roots are:±1,±2,±3,±6,±9,±18

With x=−2,

X^3−2x^2+x+18=(−2)^3−2(−2)^2+(−2)+18

X^3−2x^2+x+18=−8−8−2+18

X^3−2x^2+x+18=0

 x=−2 is a root and (x+2) a factor:

x^3−2x^2+x+18=(x+2)(x^2−4x+9)

The remaining quadratic has no real zeros, which we can decide by examining its discriminant:

X^2−4x+9

a=1,b=−4 and c=9

Δ=b^2−4ac=(−4)^2−4(1)(9)=16−36=−20

Since Δ<0 there are no real zeros and no linear factors with real (let alone rational) coefficient:

0=x^2−4x+9

0=x^2−4x+4+5

0=(x−2)^2−(√5i)2

0=(x−2−√5i)(x−2+√5i)

The roots are x=2±√5i