Rational Root Theorem

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 $f(x) = p_n x^n + p_{n-1} x^{n-1} + \cdots + p_1 x + p_0$ has a rational root of the form $r =\pm \frac {a}{b}$ with $\gcd (a,b)=1$, then $a \vert p_0$ and $b \vert p_n$.

Let's work through some examples followed by problems to try yourself.

Factorize the cubic polynomial $f(x) = 2x^3 + 7x^2 + 5x + 1$ over the rational numbers.

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By the rational root theorem, any rational root of $f(x)$ has the form $r= \frac{a}{b},$ where $a \vert 1$ and $b \vert 2$. Thus, we only need to try numbers $\pm \frac {1}{1}, \pm \frac {1}{2}$. Substituting all the possible values,

$$\begin{aligned} f(1) &> 0 \\ f(-1) &= -2 + 7 - 5 + 1 = 1 \neq 0 \\ f\bigg (\frac {1}{2}\bigg ) &> 0 \\ f\bigg (-\frac {1}{2}\bigg ) &= -\frac {2}{8} + \frac {7}{4} - \frac {5}{2} + 1 = 0. \end{aligned}$$

By the remainder-factor theorem, $(2x+1)$ is a factor of $f(x)$, implying $f(x) = (2x+1) (x^2 + 3x + 1)$. We can then use the quadratic formula to factorize the quadratic if irrational roots are desired. $_\square$

Show that $\sqrt{2}$ is irrational using the rational root theorem.

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Since $\sqrt{2}$ is a root of the polynomial $f(x) = x^2-2$, the rational root theorem states that the rational roots of $f(x)$ are of the form $\pm \frac {1,\, 2}{ 1}.$ None of these are roots of $f(x)$, and hence $f(x)$ has no rational roots. Thus, $\sqrt{2}$ is irrational.

Using this same logic, one can show that $\sqrt 3, \sqrt 5, \sqrt 7, ...$ are irrational, and from this one can prove that the square root of any number that is not a perfect square is irrational. $_\square$

Suppose $\frac {a}{b}$ is a root of $f(x)$. Then
$$p_n \left(\frac {a}{b} \right)^n + p_{n-1} \left(\frac {a}{b} \right)^{n-1} + \cdots + p_1 \frac {a}{b} + p_0 = 0.$$ By shifting the $p_0$ term to the right hand side, and multiplying throughout by $b^n$, we obtain $p_n a^n + p_{n-1} a^{n-1} b + \ldots + p_1 ab^{n-1} = -p_0 b^n$. Notice that the left hand side is a multiple of $a$, and thus $a| p_0 b^n$. Since $\gcd(a, b)=1$, Euclid's lemma implies $a | p_0$.

Similarly, if we shift the $p_n$ term to the right hand side and multiply throughout by $b^n$, we obtain $$p_{n-1} a^{n-1} b + p_{n-2} a ^{n-2} b^2 + \cdots + p_1 a b^{n-1} + p_0 b^n = -p_n a^n.$$ Note that the left hand side is a multiple of $b$, and thus $b | p_n a^n$. Since $\gcd(a,b)=1$, Euclid's lemma implies $b | p_n$. $_\square$

In particular, this tells us that if we want to check for 'nice' rational roots of a polynomial $f(x)$, we only need to check finitely many numbers of the form $\pm \frac {a}{b}$, where $a | p_0$ and $b | p_n$. 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 $f(x)$ is a monic polynomial (leading coefficient of 1), then the rational roots of $f(x)$ must be integers.

By the rational root theorem, if $r = \frac {a}{b}$ is a root of $f(x)$, then $b | p_n$. But since $p_n = 1$ by assumption, $b=1$ and thus $r=a$ is an integer. $_\square$

A short example shows the usage of the integer root theorem:

Show that if $x$ is a positive rational such that $x^2 + x$ is an integer, then $x$ must be an integer.

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Let $x^2 + x = n$, where $n$ is an integer. This is equivalent to finding the roots of $f(x) = x^2+x-n$. Since $f(x)$ is a monic polynomial, by the integer root theorem, if $x$ is a rational root of $f(x)$, then it is an integer root. $_\square$

Here is another theorem:

If $f(x)$ is a polynomial with integral coefficients, $a$ is an integral root of $f(x)$, and $m$ is any integer different from $a$, then $a-m$ divides $f(m)$.

On dividing $f(x)$ by $x-m,$ we get $f(x)=(x-m)q(x)+f(m)$, where $q(x)$ is a polynomial with integral coefficients. For $x=a$, we get $f(a)=0=(a-m)q(a)+f(m)$ or $f(m)=-(a-m)q(a)$. Hence $a-m$ divides $f(m)$. $_\square$

Let $f(x)$ be a polynomial, having integer coefficients, and let $f(0)=1989$ and $f(1)=9891$. Prove that $f(x)$ has no integer roots.

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If $a$ is an integer root of $f(x)$, then $a \neq 0$ as $f(0) \neq 0$. Also $a$ must be odd since it must divide the constant term, i.e. $f(0)=1989$. But $a \neq 1$, as $f(1) \neq 0$. So taking $m=1$ and using the above theorem, we see that the even number $(a-1)$ divides the odd number $f(1)=9891$, a contradiction.

Hence $f(x)$ has no integer roots. $_\square$

Give the following problem a try to check your understandings with these theorems:

Find all rational zeroes of $P(x) = 2x^4 + x^3 -19x^2 -9x + 9$.

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Using rational root theorem, we have the following:

• Factors of the constant term are $\pm1 , \pm3 , \pm9 .$
• Factors of the leading coefficient are $\pm1 , \pm2.$
• Possible values of rational roots are $\frac{\pm1}{1} , \frac{\pm1}{2} , \frac{\pm3}{1} , \frac{\pm3}{2} , \frac{\pm9}{1} , \frac{\pm9}{2},$ or simply $\pm1 , \frac{\pm1}{2} , \pm3 , \frac{\pm3}{2} , \pm9 , \frac{\pm9}{2}.$

Now, substituting these values in $P(x)$ and checking if it equates to zero (please refer to this: Remainder Factor Theorem), we find that $P(x) = 0$ for the values $\frac{1}{2} , 3 , -3 ,1.$

Therefore, the rational zeroes of $P(x)$ are $-3, -1, \frac{1}{2}, 3.$ $_\square$