Question

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

Answer 1

Use the discriminant to find that it has `1` irrational real zero and `2` non-real complex zeros.

Explanation:

The cubic equation:

`x^3-x^2-2x+7 = 0`

is in the standard form:

`ax^3+bx^2+cx+d = 0`

Its discriminant `Delta` is given by the formula:

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

In our example, `a=1`, `b=-1`, `c=-2` and `d=7`, so we find:

`Delta = 4+32+28-1323+252 = -1007`

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

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We were also asked if the roots were rational.

By the rational roots theorem, any rational zeros of this cubic must be expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `7` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational roots are:

`+-1, +-7`

None of these work, so our cubic has no rational roots.