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

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:

and

In rational root theorem, when $b_{n}$ = 1, it becomes the special case of

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 $b_{n}$.

Example 1:

$3a^{3}+5a^{2}+5a-2 = 0$

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

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

Solution :

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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