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

Feb 28, 2016

The FTOA allows us to infer that $F \left(x\right) = 0$ 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 $f \left(z\right)$ of degree $n > 0$ has a zero ${r}_{1} \in \mathbb{C}$ then the polynomial is divisible by $\left(z - {r}_{1}\right)$ to give a polynomial of degree $n - 1$. If $n - 1 > 0$ then this simpler polynomial will have a zero ${r}_{2} \in \mathbb{C}$ (which may be equal to ${r}_{1}$). We can then divide the simpler polynomial by $\left(z - {r}_{2}\right)$, etc. Repeat this process until you find:

$f \left(z\right) = a \left(z - {r}_{1}\right) \left(z - {r}_{2}\right) \ldots \left(z - {r}_{n}\right)$

where ${r}_{1} , {r}_{2} , \ldots , {r}_{n} \in \mathbb{C}$

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

What else can we find out about these roots of $F \left(x\right) = 0$ ?

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

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

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

Hence we find that $F \left(x\right) = 0$ 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 $\frac{p}{q}$ 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 $F \left(4\right) = 0$. So $x = 4$ is a root and $\left(x - 4\right)$ a factor:

${x}^{3} - 10 {x}^{2} + 27 x - 12 = \left(x - 4\right) \left({x}^{2} - 6 x + 3\right)$

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

$x = \frac{6 \pm \sqrt{24}}{2} = 3 \pm \sqrt{6}$

So there are $3$ positive Real roots.