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

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asked 1 week ago in STATISTICS by anonymous

1 Answer

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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
answered 4 days ago by lilly Expert
reshown 4 days ago by bradely

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