This section deals with polynomials which have integer coefficients only.
is a polynomial with integer coefficients, the polynomial
does not have only integer coefficients!
You will learn how to find all those roots of such polynomials, which are rational numbers, such as
This is not merely an esoteric exercise. Suppose, you would like to factor the polynomial
There is a general formula for the roots of a polynomial of degree 4, but it is VERY tedious to apply. It so happens that in this case x=1 and x=1 are two rational zeros (=two roots, which are rational numbers). Do you remember how to check this? To say that x=1 is a root of P(x) is the same as saying that P(1)=0.
This means that (x1)(x+1) divides evenly into
so after a polynomial long division, we are left with factoring a quadratic polynomial, which in this case turns out to have the complex roots . (Check the details!)
Suppose the polynomial has a rational root, let's call it . I will assume that p and q are coprime, i.e., the fraction is reduced to lowest terms.
What we will be doing is somewhat similar to "factoring by guessing" of quadratic polynomials.
Since is assumed to be a root of P(x), we know that :
If we multiply both sides by , we obtain:
Transfer the to the other side, and factor out a p on the left:
Now the left side is divisible by p; consequently, is divisible by p. Since p does not divide q, it does not divide , so p divides 2, i.e., or .
Start again:
This time we transfer all but the first term to the other side, and factor out a q on the right side:
Now the right side is divisible by q; consequently, the left side, the term is divisible by q. Since p and q are coprime, this means that q divides 1, i.e., .
What have we shown? Every rational root of P(x) is one of the following 4 choices: . (these are the only four numbers with p dividing 2 and q dividing 1).
We can now check each one of them:
Consequently, the polynomial has 2 rational roots: x=1 and x=1 are the only rational zeros of the polynomial P(x).
Note that in our example the leading coefficient was 1, and the constant term was 2. These were really the only two pieces of information we needed to find all rational zeros:
Suppose
is a polynomial with integer coefficients, and is a rational zero of P(x). Then

Find all rational zeros of
The leading coefficient is 6, the constant coefficient is 2. If this polynomial has rational zeros , then p divides 2 and q divides 6. Thus we have the following choices for p: ; for q our choices are: .
The candidates for rational zeros are (in decreasing order of magnitude):
Now you have to check which (if any) of these 12 values are actually roots of P(x). Doing this by hand will be tedious. Having a graphing calculator comes in handy; here is the graph of P(x) for the region we are interested in, for xvalues between 2 and 2.
There are only roots close to x=1/2 and close to x=2/3. This reduces our list of candidates to just two; plugging these values into the polynomial, we see that P(1/2)=0 and P(2/3)=0, so both are indeed rational zeros. (You have to perform this last step! It could be that P(x) does not have a rational root at 2/3, but an irrational root very close to 2/3. You would not be able to distinguish between these two cases just by looking at the graph.)
completely over the real numbers.
Let's find all rational zeros: they all have the form , where p divides the constant term 2, and q divides the leading coefficient 5. The choices for p are , the choices for q are . This leaves eight possible choices for rational zeros:
If we plug these values into the polynomial P(x), we obtain
while for the other five choices.
Consequently
divides evenly into P(x). If we perform polynomial long division, we see that
Using the quadratic formula we find, that has the real roots
Putting it all together, we obtain the following factorization for P(x):
Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.
Helmut Knaust