The rational root theorem describes a relationship between the roots of a polynomial and its coefficients. Specifically, it describes the nature of any rational roots the polynomial might possess.
The rational root theorem states that if a polynomial with integer coefficients has a rational root of the form with , then and .
Let's work through some examples followed by problems to try yourself.
Factorize the cubic polynomial over the rational numbers.
By the rational root theorem, any rational root of has the form where and . Thus, we only need to try numbers . Substituting all the possible values,
By the remainder-factor theorem, is a factor of , implying . We can then use the quadratic formula to factorize the quadratic if irrational roots are desired.
Show that is irrational using the rational root theorem.
Since is a root of the polynomial , the rational root theorem states that the rational roots of are of the form None of these are roots of , and hence has no rational roots. Thus, is irrational.
Using this same logic, one can show that are irrational, and from this one can prove that the square root of any number that is not a perfect square is irrational.
Suppose is a root of . Then
By shifting the term to the right hand side, and multiplying throughout by , we obtain . Notice that the left hand side is a multiple of , and thus . Since , Euclid's lemma implies .
Similarly, if we shift the term to the right hand side and multiply throughout by , we obtain Note that the left hand side is a multiple of , and thus . Since , Euclid's lemma implies .
In particular, this tells us that if we want to check for 'nice' rational roots of a polynomial , we only need to check finitely many numbers of the form , where and . This is a great tool for factorizing polynomials.
These are some of the associated theorems that closely follow the rational root theorem. The first one is the integer root theorem.
If is a monic polynomial (leading coefficient of 1), then the rational roots of must be integers.
By the rational root theorem, if is a root of , then . But since by assumption, and thus is an integer.
A short example shows the usage of the integer root theorem:
Show that if is a positive rational such that is an integer, then must be an integer.
Let , where is an integer. This is equivalent to finding the roots of . Since is a monic polynomial, by the integer root theorem, if is a rational root of , then it is an integer root.
Here is another theorem:
If is a polynomial with integral coefficients, is an integral root of , and is any integer different from , then divides .
On dividing by we get , where is a polynomial with integral coefficients. For , we get or . Hence divides .
Let be a polynomial, having integer coefficients, and let and . Prove that has no integer roots.
If is an integer root of , then as . Also must be odd since it must divide the constant term, i.e. . But , as . So taking and using the above theorem, we see that the even number divides the odd number , a contradiction.
Hence has no integer roots.
Give the following problem a try to check your understandings with these theorems:
Find all rational zeroes of .
Using rational root theorem, we have the following:
- Factors of the constant term are
- Factors of the leading coefficient are
- Possible values of rational roots are or simply
Now, substituting these values in and checking if it equates to zero (please refer to this: Remainder Factor Theorem), we find that for the values
Therefore, the rational zeroes of are
Brilli the Ant is playing a game with Brian Till, her best friend. They are very competitive and always want to beat each other. Today, they are going to play the quadratic game.
Brilli is going to pick 3 non-zero real numbers and Brian is going to arrange the three numbers as the coefficients of a quadratic equation:
Brilli wins the game if and only if the resulting equation has two distinct rational solutions.
Who has a winning strategy?