**Remainder Theorem** is an approach of Euclidean division of polynomials. It is also termed asÂ little BÃ©zout’s theorem. According to this theorem,Â if we divide a polynomial **P(x)** by a factor **( x – a)**; that isn’t essentially an element of the polynomial; you will find a smaller polynomial along with a remainder. This remainder that has been obtained is actually a value of **P(x) at x = a, specifically P(a)**. So basically, x -a is the divisor pf P(x) if and only if P(a) = 0.Â It is applied to factorize polynomials of each degree in a swift and elegant manner.

For example: if f(a) = a^{3}-12a^{2}-42 is divided by (a-3) then the quotient will beÂ a^{2}-9a-27 and the remainder is -123.

if we put, a-3 = 0

then a = 0

Hence, f(a) = f(3) = -123

Thus, it satisfies the remainder theorem.

## Remainder Theorem Definition

The Remainder Theorem begins with a polynomial say p(x), where “p(x)” is some polynomial p whose variable is x. Then as per theorem, dividing that polynomial p(x) by some linear factor x â€“ a, where a is just some number. Here we go through long polynomial division,Â which results in some polynomial q(x) (the “q” stands for “the quotient polynomial”) and a polynomial remainder r(x). It can be expressed as:

**p(x)/x-a = q(x) + r(x)**

**Factor Theorem**

Factor Theorem is generally applied to factorizing and finding the roots of polynomial equations. It is the reversal form of the remainder theorem. Problems are solved based on the application of synthetic division and then to check for a zero remainder.

When** p(x) = 0** then **y-x** is a factor of the polynomialÂ Or if we consider the other way, thenÂ When **y-x** is a factor of the polynomial then** p(x) **=0

### Remainder Theorem Proof

Theorem functions on an actual case that a polynomial is comprehensively dividable, at least one time by its factor in order to get a smaller polynomial and **â€˜aâ€™** remainder of zero. This acts as one of the simplest ways to determine whether the value â€˜aâ€™ is a root of the polynomial **P(x)**.

That is when we divide p(x) by x-a we obtain

p(x) = (x-a)Â·q(x) + r(x),

as we know that **Dividend = (Divisor Ã— Quotient) + Remainder**

But if r(x) is simply the constant r (remember when we divide by (x-a) the remainder is a constant)…. so we obtain the following solution, i.e

p(x) = (x-a)Â·q(x) + r

Observe what happens when we have x equal to a:

p(a) = (a-a)Â·q(a) + r

p(a) = (0)Â·q(a) + r

p(a) = r

Hence, proved.

### Steps to Divide a Polynomial by a Non-Zero Polynomial

- First, arrange the polynomials (dividend and divisor) in the decreasing order of its degree
- Divide the first term of the dividend by the first term of the divisor to produce the first term of the quotient
- Multiply the divisor by the first term of the quotient and subtract this product from the dividend, to get the remainder.
- This remainder is the dividend now and divisor will remain same
- Again repeat from the first step,Â until the degree of the new dividend is less than the degree of the divisor.

**Remainder Theorem Problems**

Consider the following example:-

**Example-** **Determine that x = 1 is a root of P(x),**

**Explanation:**

It suggests that x = 1 may be a root of P(x), and (x – 1) may be a factor of P(x)

Then if we tend to divide synthetically from P(x) by (x – 1), we will get a new smaller polynomial and a remainder of zero:

Example: Find the root of the polynomial \(x^{2}- 3x – 4\).
\( f(4) = 4^{2} – 3(4) – 4\) \( f(4) = 16 – 16 = 0\) So, Â (x-4) must be a factor of \( x^{2} – 3x – 4\)
âˆ´ p (1) = (1) By the Remainder Theorem, 2 is the remainder when \(t^{3} – 2t^{2} + t + 1\) is divided by t â€“ 1. |

**Euler Remainder Theorem**

Eulerâ€™s theoremÂ states that if n and X are two co-prime positive integers, then

X^{Ï†(n)} = 1 (mod n)

where, Ï†(n) is Euler’s function or Eulerâ€™s totient function, which is equal to;

Ï†(n) = n (1-1/a).(1-1/b).(1-1/c)

where, n is a natural number, such that n = a^{p}. b^{q} . c^{r},

here, a, b, c are prime factors of n and p, q, r are positive integers.

Example: Find the Euler totient function of 35.
\(35 = 5 \times 7\) \( Ï†(35) = 35 \left ( 1 – \frac{1}{5} \right ) . \left ( 1 – \frac{1}{7} \right ) = 24 Â \) Thus the totient function of 35 is 24. |

**Questions of the form: \(\large \mathbf{\frac{m^{a}}{n}}\)**

Example: Find the remainder when \(3^{76}\) is divided by 35.
In an example above we have already found the totient function of 35, which is equal to 24. Remainder of \(\frac{76}{Ï†(35)} = \frac{76}{24}\) = 4 Remaining power is 4, which when divided by 35 given the resultant remainder. which is, \(\frac{3^{4}}{35} = \frac{81}{35} = 11\) Thus the remainder comes out to be 11. |

This was all about the remainder theorem.

I have tried to ascertain the reminders of [ (2^n) / n ] ( read as 2 to the power n divided by n, where n is a natural number. I got accurate results for all values of n from 1 to 34. But I failed to get correct remainder while solving (2^35)/ 35 ( 2 to the power 35 divided by 35.

I am getting 23 if I take (35 = 5×7) and

32 if I take (35 = 7×5).

Whereas the actual remainder is 18 !