One way to find zeros of a polynomial is trial and error. A more efficient way is through the use of the Rational Zero Theorem.

The Rational Zero Theorem

In algebra, the Rational Zero Theorem, or Rational Root Theorem, or Rational Root Test, states a constraint on rational solutions (also known as zeros, or roots) of the polynomial equation

$a_nx^n+a_{n-1}x^{n-1}+...+a_0=0$

With integer coefficients $a_n,a_{n-1},\ldots,a_0.$

If $a_0$ and $a_n$ are nonzero, then each rational solution $x= \frac {p}{q}$, where $p$ and $q$ are coprime integers (i.e. their greatest common divisor is $1$), satisfies:

• $p$ is a divisor of the constant term $a_0$_.
• $q$ is a divisor of the leading coefficient $a_n$.

So $a_0$ must be a multiple of $p$, and $a_n $ must be a multiple of $q$.

Since any integer has only a finite number of divisors, the rational root theorem provides us with a finite number of candidates for rational roots. When given a polynomial with integer coefficients, we can plug in all of these candidates and see whether they are a zero of the given polynomial. Once we have found all the rational zeros (and counted their multiplicity, for example, by dividing using long division), we know the number of irrational and complex roots.

Since every polynomial with rational coefficients can be multiplied with an integer to become a polynomial with integer coefficients and the same zeros, the Rational Root Test can also be applied for polynomials with rational coefficients. 

Example

For example, every rational solution of the cubic equation

$3x^3-5x^2+5x-2=0$

must be among the numbers symbolically indicated by:

$\pm \frac {1,2}{1,3}$

Cubic function

The cubic function $3x^3-5x^2+5x-2$ has one real root between $0$ and $1$. We can use the Rational Root Test to see whether this root is rational. 

i.e. its numerator must divide $2$ and its denominator must divide $3$. This gives the list of possible answers

$1,-1,2,-2,\frac 13, -\frac 13, \frac 23, -\frac 23$

These root candidates can be tested, either by plugging them in directly, or by dividing and checking to see whether there is any remainder, for example using long division. The advantage of this is that once we have found a root, we immediately have found the smaller degree polynomial of which we again wish to find the roots and the rational root theorem will provide us with even fewer candidates for this root. Moreover, once we have established a root, we must use division anyway to check whether it is a multiple root.

The disadvantage is that we have to use long division more often. When there are a lot of zero candidates for a small degree polynomial, we may just want to plug in candidates and only use division when we have found a root.

In our example, we can plug in $x_0=1$ to see that it is not a root. In fact, the left hand value is equal to $1$. 

Now we use a little trick: since the constant term of $(x-x_0)^k$ equals $x_0^k$ for all positive integers $k$, we can substitute $x$ by $t+x_0$ to find a polynomial with the same leading coefficient as our original polynomial and a constant term equal to the value of the polynomial at $x_0$. In this case we substitute $x$ with $t+1$ and obtain a polynomial in $t$ with leading coefficient $3$ and constant term $1$. Thus the candidates for zeros in this polynomial in $t$ are 

$t=\pm \frac 1{1,3}$

Thus the candidates for roots of the polynomial in $x$ must be one greater than one of these candidates:

$x=1+t=2,0,\frac 43, \frac 23$

Root candidates that do not occur on both lists are ruled out. The list of rational root candidates has thus shrunk to just $x=2$ and $x=2/3$. After checking for these candidates, we see that the only rational root (with multiplicity $1)$ is $2/3$, which can also be seen in the graph above.