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# Rational Roots Theorem

Rational roots theorem plays an important role in algebra. It is also known as rational root test. This theorem applies on polynomial equations with integer coefficients and having rational roots.

A
root of a polynomial is defined as an input value which always gives a zero on substituting that value into the given polynomial.

Rational roots theorem is a nice handy tool for getting a list of first important guesses while we were trying to estimate the roots of a polynomial.

When a polynomial with integer  coefficients is given, its possible zeroes can be found by writing the factors of constant term i.e.last term with the factors of leading coefficient (first term) and form a list of rational fractions from those.

This calculation provides us a list of possible rational roots which is actually quite difficult otherwise. Since this theorem given the list of rational roots of a polynomial equation,
hence it is named as rational roots theorem.

In this article, we are going to learn about rational roots theorem in detail. So, let us go ahead and understand about this theorem and some examples based on it.

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## State and Prove

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Statement
Rational root theorem states that for a polynomial equation having integer coefficients given by the equation

$b_{n} x^{n} + b_{n-1} x^{n-1} + ... + b_{0} = 0$

If $b_{0}, b_{n} \neq 0$

then every rational solution (say x) of the polynomial equation which is written in the form x = $\frac{p}{q}$ does satisfy the following conditions:
1) The number p will be the factor of polynomial constant term $b_{0}$.
and
2) The number q will be the factor of polynomial leading coefficient $b_{n}$.

In rational root theorem, when $b_{n}$ = 1, it becomes the special case of integral root theorem.
Proof
Let us consider a polynomial P(x) = $b_{n} x^{n} + b_{n-1} x^{n-1} + ... + b_{0}$, where each $b_{0}, ...., b_{n}\ \in \mathbb{Z}$.

Let P($\frac{p}{q}$) = 0

where p and q are coprimes belonging to $\mathbb{Z}$. Then

$P$($\frac{p}{q}$) = $b_{n}$ ($\frac{p}{q}$)$^{n}$ + $b_{n-1}$ ($\frac{p}{q}$)$^{n-1}$ + ... + $b_{1}$ ($\frac{p}{q}$) + $b_{0}$ = 0.

$b_{n}$ ($\frac{p}{q}$)$^{n}$ + $b_{n-1}$ ($\frac{p}{q}$)$^{n-1}$ + ... + $b_{1}$ ($\frac{p}{q}$) + $b_{0}$ = 0.

$b_{n} p^{n} + b_{n-1} p^{n-1}q + ... + b_{1} p q^{n-1} = - b_{0} q^{n}$

$p(b_{n} p^{n-1} + b_{n-1} p^{n-2}q + ... + b_{1} q^{n-1}) = - b_{0} q^{n}$

We notice here that the integer terms inside the parentheses multiplied by p is equal to $- b_{0} q^{n}$.
Hence p must divide $- b_{0} q^{n}$.

But the fact is that p is coprime number to q. So p is also coprime to $q^{n}$.

Thus, according to Euclid's lemma, $q^{n}$ should divide its coefficient $b_{0}$.
We could shift $b _{0}$ instead, then we get

$q(b_{n-1} p^{n-1} +b_{n-2} p^{n-2} + ... + b_{0}q^{n-1}) = -b_{n}p^{n}$
Hence, similarly we may say that q divides $b_{n}$.

## How To Find Roots

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In order to find rational roots of a polynomial equation, we may use rational roots theorem. This would include the following steps:

Step 1: Arrange the given polynomial equation in the order to descending powers.

Step 2: Note down the list of all factors of constant term in the polynomial. These values are supposed to be the potential values of numerator.

Step 3: Similarly note down the list all factors of leading coefficient in the polynomial. These value would be the possible values of denominator.

Step 4:  The factors can be positive as well as negative. So consider all negative values too and cancel out if there are any duplicates.

Step 5: Make all the possible fractions out of so found numerators and denominators. This would be a possible list of all rational roots of given polynomial equation.

## Examples

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The examples of application of rational roots theorem are as follows:

Example 1:
Find the possible rational roots of the following equation
$3a^{3}+5a^{2}+5a-2 = 0$

Solution: The numerator p would be the factors of constant term 2; i.e. $\pm$1, $\pm$ 2.

The denominator q would be the factors of leading coefficient 3; i.e. $\pm$1, $\pm$ 3.

Therefore, the possible fraction will be

$\pm$ $\frac{1, 2}{1, 3}$

Thus the list of possible roots is given below :

1, -1, 2, -2, $\frac{1}{3}$, $-\frac{1}{3}$, $\frac{2}{3}$, $-\frac{2}{3}$

Example 2: Find the possible rational roots of the equation given below :

$2x^{3}+x^{4}-7x^{2}-4x+12 = 0$

Solution :
Given equation is $2x^{3}+x^{4}-7x^{2}-4x+12 = 0$

Arranging it in descending order, we get

$x^{4}+2x^{3}-7x^{2}-4x+12 = 0$

The numerator p of the rational roots would be the factors of the constant term 12; i.e. $\pm$1, $\pm$ 2,

$\pm$ 3, $\pm$ 4, $\pm$ 6, $\pm$ 12.

Similarly, the denominator q of the rational roots would be the factors of the leading coefficient 1 ; i.e. $\pm$1.

Therefore, the possible fraction will be

$\pm$ $\frac{1, 2, 3, 4, 6, 12}{1}$

Thus the list of possible roots is given below:

1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, -12
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