Question

How do you find the number of complex, real and rational roots of `x^3-6x^2-7x-12=0`?

Answer 1

This cubic has one positive irrational real root and two non-real complex roots.

Explanation:

Given:

`f(x) = x^3-6x^2-7x-12=0`

Rational root theorem

By the rational root theorem, any rational root of this polynomial is expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-12` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational roots are:

`+-1, +-2, +-3, +-4, +-6, +-12`

Descartes' Rule of Signs

The pattern of signs of the coefficients of the polynomial is `+ - - -`. With one change of signs Descartes' Rule of Signs tells us that this cubic has exactly one positive root.

We find that:

`f(6) = -54 < 0`

`f(12) = 768 > 0`

So the real positive root cannot be rational and must be an irrational number in `(6, 12)`.

Reversing the signs on the terms of odd degree, we get the pattern `- - + -`. With two changes of sign, we can deduce that this cubic has `0` or `2` negative real roots.

Discriminant

The discriminant `Delta` of a cubic polynomial in the form `ax^3+bx^2+cx+d` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

In our example, `a=1`, `b=-6`, `c=-7` and `d=-12`, so we find:

`Delta = 1764+1372-10368-3888-9072 = -20192`

Since `Delta < 0` this cubic has `1` Real zero and `2` non-Real Complex zeros, which are Complex conjugates of one another.