Question

Could someone State the possible number of imaginary zeros of `f(x)= 2x^3 - 4x^2 + 8x - 3`?

Answer 1

Descartes' Rule of Signs allows us to determine that `f(x)` has `0` or `2` non-real complex zeros. The cubic discriminant allows us to find that there are `2`.

Explanation:

Given:

`f(x) = 2x^3-4x^2+8x-3`

Fundamental Theorem of Algebra

This rather misnamed theorem (being neither fundamental nor a theorem of algebra) tells us that any non-zero polynomial with complex (possibly real) coefficients in a single variable has a complex (possibly real) zero.

As a consequence, any polynomial of degree `n > 0` in a single variable with complex (possibly real) coefficients has exactly `n` zeros counting multiplicity.

So in our example, `f(x)`, being of degree `3`, has exactly `3` complex (possibly real) zeros counting multiplicity.

Descartes' Rule of Signs

Note that the pattern of the signs of the coefficients of `f(x)` is `+ - + -`. With `3` changes of sign, Descartes' Rule of Signs tells us that `f(x)` has `1` or `3` positive real zeros.

The pattern of signs of the coefficients of `f(-x)` is `- - - -`. With no changes of sign, we can tell that `f(x)` has no negative real zeros.

So using Descartes' Rule of Signs, all we can tell is that `f(x)` has `0` or `2` non-real complex zeros.

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=2`, `b=-4`, `c=8` and `d=-3`, so we find:

`Delta = 1024-4096-768-972+3456 = -1356`

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

Derivative

`f'(x) = 6x^2-8x+8`

`color(white)(f'(x)) = 2/3(9x^2-12x+12)`

`color(white)(f'(x)) = 2/3((3x)^2-2(3x)(2)+2^2+8)`

`color(white)(f'(x)) = 2/3((3x-2)^2+8) >= 16/3" "` for all `x in RR`

So `f(x)` is strictly monotonically increasing for all real values of `x`. So it has just one real zero and two non-real complex ones.