Rational Zeros of Polynomials

This section deals with polynomials which have integer coefficients only.

displaymath151

is a polynomial with integer coefficients, the polynomial

displaymath152

does not have only integer coefficients!

You will learn how to find all those roots of such polynomials, which are rational numbers, such as

displaymath153

This is not merely an esoteric exercise. Suppose, you would like to factor the polynomial

displaymath154

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 (x-1)(x+1) divides evenly into

displaymath155

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 tex2html_wrap_inline173 . (Check the details!)


An introductory example.

How did I know that x=1 and x=-1 were roots of the polynomial

displaymath175

Suppose the polynomial has a rational root, let's call it tex2html_wrap_inline191 . 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 tex2html_wrap_inline197 is assumed to be a root of P(x), we know that tex2html_wrap_inline201 :

displaymath176

If we multiply both sides by tex2html_wrap_inline203 , we obtain:

displaymath177

Transfer the tex2html_wrap_inline205 to the other side, and factor out a p on the left:

displaymath178

Now the left side is divisible by p; consequently, tex2html_wrap_inline211 is divisible by p. Since p does not divide q, it does not divide tex2html_wrap_inline203 , so p divides 2, i.e., tex2html_wrap_inline223 or tex2html_wrap_inline225 .

Start again:

displaymath177

This time we transfer all but the first term to the other side, and factor out a q on the right side:

displaymath180

Now the right side is divisible by q; consequently, the left side, the term tex2html_wrap_inline231 is divisible by q. Since p and q are coprime, this means that q divides 1, i.e., tex2html_wrap_inline241 .

What have we shown? Every rational root tex2html_wrap_inline191 of P(x) is one of the following 4 choices: tex2html_wrap_inline247 . (these are the only four numbers with p dividing 2 and q dividing 1).

We can now check each one of them:

eqnarray46

Consequently, the polynomial has 2 rational roots: x=1 and x=-1 are the only rational zeros of the polynomial P(x).


The Rational Zero Test.

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

displaymath259

is a polynomial with integer coefficients, and tex2html_wrap_inline197 is a rational zero of P(x). Then

displaymath260


Another example.

Find all rational zeros of

displaymath269

The leading coefficient is 6, the constant coefficient is -2. If this polynomial has rational zeros tex2html_wrap_inline197 , then p divides -2 and q divides 6. Thus we have the following choices for p: tex2html_wrap_inline285 ; for q our choices are: tex2html_wrap_inline289 .

The candidates for rational zeros are (in decreasing order of magnitude):

displaymath270

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 x-values 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.)


Example 3.

Factor the polynomial

displaymath315

completely over the real numbers.

Let's find all rational zeros: they all have the form tex2html_wrap_inline197 , where p divides the constant term -2, and q divides the leading coefficient 5. The choices for p are tex2html_wrap_inline247 , the choices for q are tex2html_wrap_inline345 . This leaves eight possible choices for rational zeros:

displaymath316

If we plug these values into the polynomial P(x), we obtain

displaymath317

while tex2html_wrap_inline349 for the other five choices.

Consequently

displaymath318

divides evenly into P(x). If we perform polynomial long division, we see that

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Using the quadratic formula we find, that tex2html_wrap_inline353 has the real roots

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Putting it all together, we obtain the following factorization for P(x):

displaymath321


Exercise 1.

Find all rational zeros of the polynomial

displaymath357

Answer.

Exercise 2.

Find all rational zeros of the polynomial

displaymath359

Answer.

Exercise 3.

Find all rational zeros of the polynomial

displaymath361

Answer.

Exercise 4.

Factor the polynomial in Exercise 3 completely (a) over the real numbers, (b) over the complex numbers.

Answer.

Exercise 5.

Show the following: If a polynomial has integer coefficients and its leading coefficient is 1, then all of its rational zeros are in fact integers.

Answer.

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