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 \(-3, -1, \frac{1}{2}, 3.\) \(_\square\)