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Explore the Back-StoryIn this post, we will be learning about the **Rational Root Theorem Proof**. It is a great tool from Algebra and is useful for the Math Olympiad Exams and ISI and CMI Entrance Exams.

So, here is the starting point....

$a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{2} x^{2}+a_{1} x+a_{0}$

This polynomial has certain properties.

*1. The coefficients are all integers. That means, $a_{n}$, $a_{n-1}$, $\ldots a_{0}$ are all integers.*

*2. $a_{n}$, $a_{0} \neq 0$.*

Now, we are interested to find out if this equation:

$a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{2} x^{2}+a_{1} x+a_{0}$ = 0 has a rational root.

So, here is a question:

Is there a rational number p over q that satisfies this equation?

Before we even go into solving this one or rather showing you what the rational root theorem is all about, think about the power of this theorem.

Because there are literally infinitely many numbers which are called rational numbers which are of the form p over q where p is an integer and q is an integer and we often assume that the HCF of p and q is 1. So, you are commenting on infinity in some sense. You are saying something about infinitely many numbers. That's always powerful!

**Now, let's learn it step by step: **

If p/q is a solution or root, then let's plugin p/q in the equation $a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}$ = 0

Now, we get:

$a_{n}\left(\frac{p}{q}\right)^{n}+a_{n-1}\left(\frac{p}{q}\right)^{n-1}+\ldots+a_{1}\left(\frac{p}{q}\right)+a_{0}$ = 0

Multiplying both sides by $q^{n}$,

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

Moving $a_{0} q^{n}$ to R.H.S,

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

Taking p common from the L.H.S, we have:

$p\left(a_{n} p^{n-1}+a_{n-1} p^{n-2} q+\ldots+a_{1} q^{n-1}\right)$ = $-a_{0} q^{n}$

Notice, that the L.H.S is divisible by p, so, we understand that the R.H.S is also divisible by p.

That means p divides $-a_{0} q^{n}$

Now, remember that the H.C.F p over q is 1. So, p cannot divide $q^{n}$.

That means p divides $a_{0}$.

So, similarly, we can prove that q divides $a_{n}$. (Try this out yourself)

**Caution:** If p divides $a_{0}$ and q divides $a_{n}$, that does not automatically mean p/q is a root.

So, you have to plug in and check and if you have checked all possible combinations of them and none of them actually produced a zero, then it's very simple. It does not have a rational root.

We hope the Rational Root Theorem Proof is now clear to you. You can also watch the video tutorial from the link given below.

In this post, we will be learning about the **Rational Root Theorem Proof**. It is a great tool from Algebra and is useful for the Math Olympiad Exams and ISI and CMI Entrance Exams.

So, here is the starting point....

$a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{2} x^{2}+a_{1} x+a_{0}$

This polynomial has certain properties.

*1. The coefficients are all integers. That means, $a_{n}$, $a_{n-1}$, $\ldots a_{0}$ are all integers.*

*2. $a_{n}$, $a_{0} \neq 0$.*

Now, we are interested to find out if this equation:

$a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{2} x^{2}+a_{1} x+a_{0}$ = 0 has a rational root.

So, here is a question:

Is there a rational number p over q that satisfies this equation?

Before we even go into solving this one or rather showing you what the rational root theorem is all about, think about the power of this theorem.

Because there are literally infinitely many numbers which are called rational numbers which are of the form p over q where p is an integer and q is an integer and we often assume that the HCF of p and q is 1. So, you are commenting on infinity in some sense. You are saying something about infinitely many numbers. That's always powerful!

**Now, let's learn it step by step: **

If p/q is a solution or root, then let's plugin p/q in the equation $a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}$ = 0

Now, we get:

$a_{n}\left(\frac{p}{q}\right)^{n}+a_{n-1}\left(\frac{p}{q}\right)^{n-1}+\ldots+a_{1}\left(\frac{p}{q}\right)+a_{0}$ = 0

Multiplying both sides by $q^{n}$,

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

Moving $a_{0} q^{n}$ to R.H.S,

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

Taking p common from the L.H.S, we have:

$p\left(a_{n} p^{n-1}+a_{n-1} p^{n-2} q+\ldots+a_{1} q^{n-1}\right)$ = $-a_{0} q^{n}$

Notice, that the L.H.S is divisible by p, so, we understand that the R.H.S is also divisible by p.

That means p divides $-a_{0} q^{n}$

Now, remember that the H.C.F p over q is 1. So, p cannot divide $q^{n}$.

That means p divides $a_{0}$.

So, similarly, we can prove that q divides $a_{n}$. (Try this out yourself)

**Caution:** If p divides $a_{0}$ and q divides $a_{n}$, that does not automatically mean p/q is a root.

So, you have to plug in and check and if you have checked all possible combinations of them and none of them actually produced a zero, then it's very simple. It does not have a rational root.

We hope the Rational Root Theorem Proof is now clear to you. You can also watch the video tutorial from the link given below.

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

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