Rational Root Theorem

To prove this theorem, each of the properties will be considered individually. To start with the first property, consider a polynomial

P (x)

with integer coefficients that has an integer root

x_{r} .

This means that

P (x_{r}) = 0 .

a_{n} x_{r}^{n} + a_{n - 1} x_{r}^{n - 1} + \dots + a_{1} x_{r} + a_{0} = 0

Now

a_{0}

is subtracted from both sides of the equation.

a_{n} x_{r}^{n} + a_{n - 1} x_{r}^{n - 1} + \dots + a_{1} x_{r} + a_{0} = 0

SubEqn

$LHS - a_{0} = RHS - a_{0}$

a_{n} x_{r}^{n} + a_{n - 1} x_{r}^{n - 1} + \dots + a_{1} x_{r} = - a_{0}

FactorOut

Factor out $x_{r}$

x_{r} (a_{n} x_{r}^{n - 1} + a_{n - 1} x_{r}^{n - 2} + \dots + a_{1}) = - a_{0}

Since the coefficients of

P (x)

are integers and

x_{r}

is an integer, the expression between the parentheses results in an integer. Also,

- a_{0}

is the product of an integer and

x_{r},

which means that

x_{r}

is a factor of

a_{0} .

This completes the proof of the first property.

For the second property, suppose that the polynomial $P (x)$ has a rational root $\frac{p}{q},$ such that the fraction is written in its simplest form. Again, it is obtained that $P (\frac{p}{q}) = 0 .$

$a_n \left(\dfrac{p}{q}\right)^n + a_{n-1}\left(\dfrac{p}{q}\right)^{n-1}+\cdots+a_1\left(\dfrac{p}{q}\right)+a_0 = 0$

This time the equation will be modified in a different way.

a_{n} (\frac{p}{q})^{n} + a_{n - 1} (\frac{p}{q})^{n - 1} + \dots + a_{1} (\frac{p}{q}) + a_{0} = 0

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

a_{n} (\frac{p ^{n}}{q ^{n}}) + a_{n - 1} (\frac{p ^{n - 1}}{q ^{n - 1}}) + \dots + a_{1} (\frac{p}{q}) + a_{0} = 0

MultEqn

$LHS \cdot q^{n} = RHS \cdot q^{n}$

a_{n} (p^{n}) + a_{n - 1} (p^{n - 1} q) + \dots + a_{1} (p q^{n - 1}) + a_{0} (q^{n}) = 0

SubEqn

$LHS - a_{0} (q^{n}) = RHS - a_{0} (q^{n})$

a_{n} (p^{n}) + a_{n - 1} (p^{n - 1} q) + \dots + a_{1} (p q^{n - 1}) = - a_{0} (q^{n})

FactorOut

Factor out $p$

p (a_{n} p^{n - 1} + a_{n - 1} p^{n - 2} q + \dots + a_{1} q^{n - 1}) = - a_{0} q^{n}

Since the coefficients of

P (x),

p,

and

q

are all integers, the expression between parentheses results in an integer. Also, since

\frac{p}{q}

is written in its simplest form,

p

and

q

do not have a common factor, which means that

p

and

q^{n}

do not have a common factor either. Therefore,

p

is a factor of

a_{0} .

The numerator of the root, $p,$ is a factor of $a_{0} .$

If the modification is done differently, it is possible to come to the other property.

a_{n} (p^{n}) + a_{n - 1} (p^{n - 1} q) + \dots + a_{1} (p q^{n - 1}) + a_{0} (q^{n}) = 0

SubEqn

$LHS - a_{n} p^{n} = RHS - a_{n} p^{n}$

a_{n - 1} p^{n - 1} q + \dots + a_{1} p q^{n - 1} + a_{0} q^{n} = - a_{n} p^{n}

FactorOut

Factor out $q$

q (a_{n - 1} p^{n - 1} + \dots + a_{1} p q^{n - 2} + a_{0} q^{n - 1}) = - a_{n} p^{n}

With the same reasoning as before, it is possible to conclude that

q,

the denominator of the root, is a factor of

a_{n},

finishing the proof.

Rational Root Theorem

Proof

Recommended exercises