A Remainder Theorem is an approach of Euclidean division of polynomials. â€œIt is applied to factorize polynomials of each degree in swift and elegant manner. Theorem implies that after you 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)**Â â€œ.

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 way 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

**Remainder Theorem Examples**

Consider the following example:-

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

**Explanation:**

It suggest 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:

**Factor Theorem**

Factor Theorem is generally applied to factorizing and finding the roots of polynomial equations. It is the reversal form of 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

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

**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, \(Ï†(n)\)

where Ï†(n) is Eulerâ€™s totient function and

\( Ï†(n) = n \left ( 1- \frac{1}{a} \right ) . \left ( 1- \frac{1}{b} \right ) . \left ( 1- \frac{1}{c} \right )\)

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.

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

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

In an example above we have already found the totient function of 35, which is equal to 24. Remainder of \(\frac{76}{Ï†(35)} = \frac{76}{24}\) 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. |

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