Question

How do you find the number of roots for `f(x) = 4x^3 – x^2 – 2x + 1` using the fundamental theorem of algebra?

Answer 1

It has `3` zeros since it is of degree `3`.

Explanation:

The fundamental theorem of algebra (FTOA) tells you that a polynomial of non-zero degree has at least one Complex (possibly Real) zero.

A straightforward corollary of this - often stated as part of the FTOA - is that a polynomial in one variable of degree `n > 0` has exactly `n` Complex (possibly Real) zeros counting multiplicity.

To see this, note that if `f(x)` is of degree `n > 0`, then by the FTOA it has a zero `r_1 in CC`. Then `(x-r_1)` is a factor of `f(x)` and `f(x)/(x-r_1)` is a polynomial of degree `n - 1`. If `n-1 > 0` then `f(x)/(x-r_1)` has a zero `r_2 in CC` (possibly equal to `r_1`), so a factor `(x-r_2)`, etc. Repeat to find all `n` zeros.

In our example `f(x)` is of degree `3` so has exactly `3` zeros counting multiplicity.

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Bonus

For more information about the zeros, we can evaluate the discriminant of the cubic.

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 this example, `a=4`, `b=-1`, `c=-2`, `d=1` and we find:

`Delta = 4+128+4-432+144 = -152 < 0`

Since `Delta < 0` this cubic has one Real zero and a Complex conjugate pair of non-Real zeros.

We can use Descartes' rule of signs to tell that `f(x)` has `0` or `2` positive Real zeros and exactly one `1` negative zero. Since we already know that it only has one Real zero, it is negative.