Rational Root Theorem – Proof

15 Feb
Mathematics Theorems, , , , , ,
Ex: 8.1, Class 10
Ex: 8.3, Class 10

Rational Root Theorem

Statement: Given a polynomial

𝑃(𝑥)=𝑎𝑛𝑥𝑛+𝑎𝑛−1𝑥𝑛−1+𝑎𝑛−2𝑥𝑛−2+…….+𝑎1𝑥+𝑎0

with integral coefficients where an ≠ 0, a0 ≠ 0. if P(x) has a rational root p/q where p and q are relatively prime integers, then p is a divisor of a0 and q is a divisor of an.

Proof

𝑃(𝑥)=𝑎𝑛𝑥𝑛+𝑎𝑛−1𝑥𝑛−1+𝑎𝑛−2𝑥𝑛−2+…….+𝑎1𝑥+𝑎0𝑤ℎ𝑒𝑟𝑒  𝑎0,…,𝑎𝑛∈ℤ  𝑎𝑛𝑑  𝑎0,𝑎𝑛≠0.

Let p/q be a rational root of this equation where p and q are coprime integers and q ≠ 0.

⇒𝑃(𝑝𝑞)=0⇒𝑎𝑛(𝑝𝑞)𝑛+𝑎𝑛−1(𝑝𝑞)𝑛−1+𝑎𝑛−2(𝑝𝑞)𝑛−2+…….+𝑎1(𝑝𝑞)+𝑎0=0

Multiply both sides by qn

𝑞𝑛×[𝑎𝑛(𝑝𝑞)𝑛+𝑎𝑛−1(𝑝𝑞)𝑛−1+𝑎𝑛−2(𝑝𝑞)𝑛−2+…….+𝑎2(𝑝𝑞)2+𝑎1(𝑝𝑞)+𝑎0]=𝑞𝑛×0𝑎𝑛(𝑝𝑞)𝑛𝑞𝑛+𝑎𝑛−1(𝑝𝑞)𝑛−1𝑞𝑛+𝑎𝑛−2(𝑝𝑞)𝑛−2𝑞𝑛+…….+𝑎2(𝑝𝑞)2𝑞𝑛+𝑎1(𝑝𝑞)𝑞𝑛+𝑎0𝑞𝑛=0𝑎𝑛𝑝𝑛+𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1+𝑎0𝑞𝑛=0  …(1)

Imp: We multiplied by qn to get rid of terms in the denominator. 

Video tutorial: Proof of Rational Root Theorem 

Proof of Part 1: p divides a0

Let’s write equation (1) again

𝑎𝑛𝑝𝑛+𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1+𝑎0𝑞𝑛=0  …(1)

Take a0qn to the other side of the equation.

𝑎𝑛𝑝𝑛+𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1=  –𝑎0𝑞𝑛

Now p is common in every term on LHS. This implies: 

𝑝[𝑎𝑛𝑝𝑛−1+𝑎𝑛−1𝑝𝑛−2𝑞+𝑎𝑛−2𝑝𝑛−3𝑞2+…….+𝑎2𝑝𝑞𝑛−2+𝑎1𝑞𝑛−1]=  –𝑎0𝑞𝑛

Now p, q and a0, a1,….an, all are integers. Hence LHS is an integer.

⇒𝑝|𝐿𝐻𝑆⇒𝑝|𝑅𝐻𝑆⇒𝑝|𝑎0𝑞𝑛

As we know that p and q are coprime 

⇒𝑝  does not divide  𝑞⇒𝑝  does not divide  𝑞𝑛
But because  𝑝|𝑎0𝑞𝑛⇒𝑝|𝑎0

Proof of Part 2: q divides an

Let’s go back to equation (1)

𝑎𝑛𝑝𝑛+𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1+𝑎0𝑞𝑛=0  …(1)

Take anpn to the other side of the equation.

𝑎𝑛−1𝑝𝑛−1𝑞+𝑎𝑛−2𝑝𝑛−2𝑞2+…….+𝑎2𝑝2𝑞𝑛−2+𝑎1𝑝𝑞𝑛−1+𝑎0𝑞𝑛=−𝑎𝑛𝑝𝑛  

Now q is common in every term on LHS. This implies: 

𝑞[𝑎𝑛−1𝑝𝑛−1+𝑎𝑛−2𝑝𝑛−2𝑞+…….+𝑎2𝑝2𝑞𝑛−3+𝑎1𝑝𝑞𝑛−2+𝑎0𝑞𝑛−1]=−𝑎𝑛𝑝𝑛

Now p, q and a0, a1,….an, all are integers. Hence LHS is an integer.

⇒𝑞|𝐿𝐻𝑆⇒𝑞|𝑅𝐻𝑆⇒𝑞|𝑎𝑛𝑝𝑛

As we know that p and q are coprime 

⇒𝑞  does not divide  𝑝⇒𝑞  does not divide  𝑝𝑛
But because  𝑞|𝑎𝑛𝑝𝑛⇒𝑞|𝑎𝑛

Hope you find this proof useful.

Let’s now use this theorem to find the roots of a polynomial of degree 3. 

Example of Rational Root Theorem

𝑃(𝑥)=2𝑥3–3𝑥2–2𝑥+3

Find the roots of P(x) using Rational Root Theorem.

This equation is of the form

𝑃(𝑥)=𝑎3𝑥3+𝑎2𝑥2+𝑎1𝑥+𝑎0
⇒𝑎3=2  𝑎𝑛𝑑  𝑎0=3

As per rational root theorem, if this polynomial has any rational root, say p/q, then

⇒  𝑝  𝑑𝑖𝑣𝑖𝑑𝑒𝑠  𝑎0  ,  𝑞  𝑑𝑖𝑣𝑖𝑑𝑒𝑠  𝑎3𝑝  𝑑𝑖𝑣𝑖𝑑𝑒𝑠  3  ,  𝑞  𝑑𝑖𝑣𝑖𝑑𝑒𝑠  2

Here are the possible values of p/q:

𝑝𝑞⇒𝑝𝑞=±1,  ±3=±1,  ±2=±11,±31,±12,±32=±1,±3,±12,±32

Now, if P(x) has any rational root, it has to be from the possible values of p/q written above. If P(x) = 0 for any of these values, that will be a rational root of the polynomial. The three values for which P(x) = 0 are 1, -1 and 3/2.

𝑃(𝑥)𝑃(1)𝑃(−1)𝑃(32)=2𝑥3–3𝑥2–2𝑥+3=2−3−2+3=0=−2−3+2+3=0=2×278−3×94−2×32+3=274–274−3+3=0

Rational Root Theorem is a very important theorem of Polynomials and is intended for the students learning higher mathematics and for those who are willing to dive deep into the core mathematics. For lower classes, like class 9 and class 10, this theorem comes handy when we have to find the roots of polynomials with degree 3 or more. 

Ex: 8.1, Class 10
Ex: 8.3, Class 10
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