Question

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

Answer 1

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`