Question

How do you find the number of roots for `F(x)=x^3-10x^2+27x-12` using the fundamental theorem of algebra?

Answer 1

The FTOA allows us to infer that `F(x) = 0` has `3` roots.

Further analysis allows us to find the `3` Real roots that it has.

Explanation:

The fundamental theorem of algebra tells you that any non-constant polynomial with Complex (possibly Real) coefficients has a zero in the Complex numbers.

A simple corollary of this is that a polynomial of degree `n > 0` will have `n` zeros in the Complex numbers, counting multiplicity.

To show the corollary, notice that if a polynomial `f(z)` of degree `n > 0` has a zero `r_1 in CC` then the polynomial is divisible by `(z-r_1)` to give a polynomial of degree `n-1`. If `n-1 > 0` then this simpler polynomial will have a zero `r_2 in CC` (which may be equal to `r_1`). We can then divide the simpler polynomial by `(z-r_2)`, etc. Repeat this process until you find:

`f(z) = a(z-r_1)(z-r_2)...(z-r_n)`

where `r_1, r_2,..., r_n in CC`

In our example `F(x)` is of degree `3`, so has exactly `3` Complex (possibly Real) zeros counting multiplicity.

What else can we find out about these roots of `F(x) = 0` ?

Notice that the coefficients of `F(x)` have `3` changes of sign. So `F(x)` may have up to `3` positive Real zeros.

On the other hand, `F(-x)` has no changes of sign, so `F(x)` has no negative Real zeros.

Note that since `F(x)` has Real coefficients, any non-Real Complex roots must occur in Complex conjugate pairs.

Hence we find that `F(x) = 0` either has `3` positive Real roots or `1` positive Real roots and a pair of non-Real Complex conjugate roots.

By the rational root theorem, any rational roots must be 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`

(positive since we know there are no negative zeros).

Trying each of these in turn we find `F(4) = 0`. So `x=4` is a root and `(x-4)` a factor:

`x^3-10x^2+27x-12 = (x-4)(x^2-6x+3)`

We can solve the remaining quadratic factor using the quadratic formula to find roots:

`x = (6+-sqrt(24))/2 = 3+-sqrt(6)`

So there are `3` positive Real roots.