Question

What are the possible rational roots of `x^3-4x^2+x+2=0` and then determine the rational roots?

Answer 1

"Possible" rational roots: `+-1`, `+-2`

Rational root: `1`

Irrational roots: `3/2+-sqrt(17)/2`

Explanation:

Given:

`x^3-4x^2+x+2 = 0`

By the rational roots theorem, any rational roots of this cubic are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `2` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational roots are:

`+-1`, `+-2`

In addition, notice that the sum of the coefficients is zero. That is:

`1-4+1+2 = 0`

Hence we can tell that `x=1` is a zero and `(x-1)` a factor:

`0 = x^3-4x^2+x+2`

`color(white)(0) = (x-1)(x^2-3x-2)`

The remaining roots can be found using the quadratic formula.

`x = (3+-sqrt((-3)^2-4(1)(-2)))/(2*1) = 3/2+-sqrt(17)/2`