Rational Root Theorem

The Rational Root Theorem $\text{(RRT)}$ is a method that provides a quick test for the rationality of some expressions. It also helps to find rational roots of polynomials. The Rational Root Theorem states that if a rational root of a polynomial exists, then its components will divide the first and last coefficients. In this article, we will learn different roots of polynomials such as rational and integral roots, and how to find the roots of a polynomial along with the solved example.

Rational Root Theorem Explanation

Consider a polynomial of degree n, having integer coefficients:

$a_0nx^n + a_{n-1}x^{(n-1)} + … + a_2x^2 + a_1x + a_0 = 0$

Rational Root $\dfrac{p}{q}$ GCD$(p,q) = 1$

The Rational Root Is Expressed in the Lowest Terms it Can Be in Rational Form. it Means P and Q Have No Common Factors. the Numerator Divides the Constant at the End of the Polynomial and the Denominator Divides the Leading coefficient.

Consider an example:

$2x^3 - 7x^2 + 3x + 12 = 0$

Here we will only need  $2$ and the $12$. If a rational root $\dfrac{p}{q}$ exists, then:

The factors of $12$ are: $\pm 1,2,3,4,6,12$

The factors of $2: \pm 1,2$

Therefore, if a rational root does exist, it’s one of these:

$\pm 1,2,3,4,6,12 \\ \pm \dfrac{1}{2}, \dfrac{3}{2}$

Integral Roots

Any number that satisfies the value of a polynomial is called its roots. The roots which are integers i.e they are neither irrational nor imaginary are called integral roots. 

How to Find the Roots of a Polynomial?

Roots of polynomials are defined as the solutions for any given polynomial for which we need to find the value of the unknown variable. If we know the roots of a polynomial, we can evaluate the value of the polynomial to zero. An expression which is in the form $a_nx^n + a_{(n-1)}x^{(n-1)} + … + a_1x + a_0$, where each variable has a constant accompanying it as its coefficient is called a polynomial of degree $n$ in variable $x$. Each variable separated with an addition or subtraction symbol in the expression is known as the term. The degree of a polynomial is defined as the maximum power of the variable of that polynomial.

For example, a linear polynomial of the form $ax + b$ is called a polynomial that has degree $1$. Similarly, quadratic and cubic polynomials have a degree of $2$ and $3$ respectively.

A polynomial that consists of only one term is known as a monomial. A monomial that contains only a constant term is said to be a polynomial of zero degrees. A polynomial can account for a null value even if the values of the constants are greater than zero. In such cases, we look for the value of variables that set the value of the entire polynomial to zero. These values of a variable are known as the roots of polynomials. Sometimes they are also known as zeros of the polynomial.

Roots of Polynomials Formula

Consider the polynomials expression are written in the form of:

$a_nx^n + a_{(n-1)}x^{(n-1)} + … + a_1x + a_0$

The root of linear polynomial such as $ax + b$ can be calculated by the formula $x = - \dfrac{b}{a}$

The general form of a quadratic polynomial is $ax^2 + bx + c$ and if we equate this expression to zero, we get a quadratic equation, i.e. $ax^2 + bx + c = 0$

The root of the quadratic equation whose degree is two, such as $ax^2 + bx + c$ are evaluated using the formula;

$ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The formulas for higher degree polynomials are a little complicated.

Roots of Three-degree Polynomial

To find the roots of the three-degree polynomial first we need to factorise the given polynomial equation so that we will get a linear and quadratic equation. After that, we can easily determine the zeros of the three-degree polynomial. Let us understand with an example.

Example: $2x^3 - x^2 - 7x + 2$ 

Divide the above polynomial by $x – 2$ because it is one of the factors.

$2x^3 - x^2 - 7x + 2 = (x - 2)(2x^2 + 3x - 1)$

Now we can find the roots of the above polynomial since we have got one linear equation and one quadratic equation according to the known formula.

Finding Roots of Polynomials

Given $p(x) = 5x + 1$ is an example of the polynomial $p(x)$ of degree $1$:

As we know, if  $a$ is the root of a polynomial $p(x)$, then  $P(a) = 0$.

To determine the roots of polynomials $p(x)$, we have to find the value of $x$ where $p(x) = 0$. Now,

$5x + 1 = 0$

$\Rightarrow x = - \dfrac{1}{5}$

Therefore, $- \dfrac{1}{5}$ is the root of the polynomial $p(x)$.

Constant Term of a Polynomial

The constant term of a polynomial is the term having degree 0. It is the term that is independent of the variable. 

Consider polynomial $2x^2 + 3x + 5$

Here the last term doesn't have any variable at all Because there is no variable in this last term, its value never changes, so it is called the "constant" term.

Ex- The constant term of this polynomial $5x^3 - 4x^2 + 7x - 8$ is $−8$.

Factor Theorem Definition

According to the factor theorem, if $f(x)$ is a polynomial of degree $n \geq 1$ and $a$’ is any real number, then, $(x-a)$ is a factor of $f(x)$, if $ f(a) = 0$.

Also, we can say, if $(x-a)$ is a factor of polynomial $f(x)$, then $f(a) = 0$.

We use the factor theorem for factoring a polynomial and finding the roots of the polynomial. It is also known as a special case of a polynomial remainder theorem.

As we know, a polynomial $f(x)$ has a factor $(x - a)$, if and only if, $f(a) = 0$. It is one of the methods to do the factorisation of the polynomial.

How to Use Factor Theorem 

Following are the step to find the factors of a polynomial $f(x)$ using the factor theorem:

• Step 1: If $f(-c) = 0$, then $(x + c)$ will be a factor of the polynomial $f(x)$.

• Step 2: If $p\left(\dfrac{d}{c}\right) = 0$, then $(cx - d)$ will be a factor of the polynomial $f(x)$.

• Step 3: If $p\left( - \dfrac{d}{c}\right) = 0$, then $(cx + d)$ will be a factor of the polynomial $f(x)$.

• Step 4: If $p(c)=0$ and $p(d) =0$, then $(x - c)$ and $(x - d)$ will be the factors of the polynomial $p(x)$.

Rational Roots

In Algebra, the rational root theorem, also called the rational root test. For a polynomial equation in the variable with integer coefficients to have a solution that is a rational number. The leading coefficient (or the coefficient of the highest power) should be divisible by the denominator of the fraction and the constant term (the one without a variable) should be divisible by the numerator. In algebra the canonical form for a polynomial equation in one variable $(x)$ is

$a_nx^n + a_{(n-1)}x^{(n-1)} + … + a_1x + a_0$

Where $a_0, a_1 …,$ and are ordinary integers. Thus, for a polynomial equation to have a rational solution $\dfrac{p}{q}, q$ must divide $a_n$ and $p $ must divide $a_0$.

The Rational Root Theorem important methods to know which roots we may find exactly (the rational ones) and which roots we may only approximate (the irrational ones).

Consider the Polynomial

$P(x) = x^3 - 8x^2 + 17x - 10$

In this case, $a_0 = - 10$ and $a_n = 1$. The number $ –10$  has factors $\pm  \{10, 5, 2, 1\}$. The number  $1$  has factors of   $1$. Thus, if polynomial $P$ contains any rational roots, then they must be among $\pm \begin{Bmatrix} \dfrac{10}{1}, \dfrac{5}{1}, \dfrac{2}{1}, \dfrac{1}{1} \end{Bmatrix} =  \pm \{10, 5, 2, 1\}$. Therefore  $5, 2,$ and  $1$  are all rational (actually integer) roots.

Solved Examples

1. Find the Roots of the Polynomial $x^2 + 2x - 15$.

Sol: Given polynomial $x^2 + 2x - 15$.

We will find roots of polynomial By splitting the middle term,

$x^2 + 2x - 15$

$= x(x + 5) – 3(x + 5) \\ = (x – 3) (x + 5) \\  x = 3 \text{ or } x =−5$

Hence roots of polynomials are $3$ and $-5$.

2. State the Possible Rational Zeros for Each Function. Then Find All Rational Zeros

$f (x) = 4x^3 - 9x^2 + 6x − 1$

Sol: Here $a_0 = 4$ and $a_n = -1$

Factors of the coefficient of $4x^3$ are $\pm 1, \pm 4$ and $\pm 2$

Factors of the constant term $(-1)$ are $\pm 1$

If there exists any rational root for the given cubic polynomial, it must be in the form of $\left( \pm \dfrac{1}{1}, \pm \dfrac{1}{4} \text{ and } \pm \dfrac{1}{2}\right)$

Let us check for $\dfrac{1}{4}$ 

$P(x) = 4x^3 - 9x^2 + 6x − 1 \\ P\left(\dfrac{1}{4}\right) = 4\left(\dfrac{1}{4}\right)^3 - 9\left(\dfrac{1}{4}\right)^2 + 6\left(\dfrac{1}{4}\right) − 1 \\ \Rightarrow 4\left(\dfrac{1}{64}\right) - 9\left(\dfrac{1}{16}\right) + \left(\dfrac{6}{4}\right) − 1 \\ = \left(\dfrac{1}{16}\right) - \left(\dfrac{9}{16}\right) + \left(\dfrac{6}{4}\right) − 1 \\ =  \dfrac{(1 - 9 + 24 - 16)}{16} \\ =  \dfrac{(25 - 25)}{16} = 0$

Hence $\left(\dfrac{1}{4}\right)$ is one of the rational roots of the given cubic polynomial. We can get other roots using synthetic division.

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We can get the other two roots by factorizing the quadratic polynomial $4x^2 - 8x + 4$.

Dividing the whole equation by $4$, we get 

$x^2 - 2x + 1 = 0 \\ x^2 - 2x + 1 = 0 \\ \Rightarrow (x - 1)(x - 1) = 0 \\ $ So, $x - 1 = 0$   or  $x - 1 = 0$

$x = 1$ and $x = 1$

Hence all three roots of the polynomial are $1, 1$ and $\dfrac{1}{4}$.