What are the possible rational roots of #x^3-2x^2+x+18=0# and then determine the rational roots?

1 Answer
May 7, 2017

Answer:

The "possible" rational roots are:

#+-1, +-2, +-3, +-6, +-9, +-18#

The actual roots are:

#-2" "# and #" "2+-sqrt(5)i#

Explanation:

Given:

#x^3-2x^2+x+18=0#

By 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#, we find:

#x^3-2x^2+x+18 = (color(blue)(-2))^2-2(color(blue)(-2))^2+(color(blue)(-2))+18#

#x^3-2x^2+x+18 = -8-8-2+18#

#x^3-2x^2+x+18 = 0#

So #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 tell by examining its discriminant:

#x^2-4x+9#

is in the form:

#ax^2+bx+c#

with #a=1#, #b=-4# and #c=9#

Its discriminant #Delta# is given by the formula:

#Delta = b^2-4ac = (color(blue)(-4))^2-4(color(blue)(1))(color(blue)(9)) = 16-36 = -20#

Since #Delta < 0# there are no real zeros and no linear factors with real (let alone rational) coefficients.

We can find the roots by completing the square:

#0 = x^2-4x+9#

#color(white)(0) = x^2-4x+4+5#

#color(white)(0) = (x-2)^2-(sqrt(5)i)^2#

#color(white)(0) = (x-2-sqrt(5)i)(x-2+sqrt(5)i)#

Hence roots:

#x = 2+-sqrt(5)i#