What are the possible rational roots of #x^3-2x^2+x+18=0# and then determine the rational roots?
1 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
That means that the only possible rational roots are:
#+-1, +-2, +-3, +-6, +-9, +-18#
With
#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^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
Its discriminant
#Delta = b^2-4ac = (color(blue)(-4))^2-4(color(blue)(1))(color(blue)(9)) = 16-36 = -20#
Since
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#