Algebraic Remainder Theorem

Sohel Sahoo : Hello Guys,

   Here you will get to know a trick to find out remainder of an algebraic epression or function when divided by a linear function of one variable as often asked in different competitive exams & questions are set on this concept;which is known as Remainder Theorem with full proof &detailed analysis.

Remainder of Algebraic Function:-

The Remainder Theorem states that if a polynomial f(x) is divided by a linear(of degree 1) function  (x-k),then the remainder is f(k) .

Proof:-

 In any Division by Euclidean Division Algorithm,

                      Dividend = Divisor x Quotient + Remainder

Let Q(x) be the quotient and R be the remainder.

      Then, f(x) = (x-k) x Q(x) + R  .......(1)

In order to get the remainder  the quotient part should be equal to zero.As Q(x) is a polynomial it can't be zero. So, equate (x-k) to zero.

          (x-k) x Q(x) = 0 

      =>  (x-k)= 0 => x = k

Putting  x = k in equation (1) we have,

    f(k) = 0 + R => R =  f(k)

COROLLARY:- 

If a polynomial f(x) is divided by a linear function (ax-k),then the remainder is f(k/a) .

Proof:- 

  In any Division by Euclidean Division Algorithm,

                      Dividend = Divisor x Quotient + Remainder

Let Q(x) be the quotient and R be the remainder.

      Then, f(x) = (ax-k) x Q(x) + R  .......(1)

In order to get the remainder  the quotient part should be equal to zero.As Q(x) is a polynomial it can't be zero. So, equate (ax-k) to zero.

          (ax-k) x Q(x) = 0 

      =>  (ax-k)= 0 => x = k/a

Putting  x = k/a in equation (1) we have,

    f(k/a) = 0 + R => R =  f(k/a)

FACTOR THEOREM:-

It is a special case of Remainder Theorem,i.e. perfectly divisible by the divisor case.

• Statement:- When f(k) = 0 ,then we can say that f(x) is exactly divisible by (x-k).

  Proof:- By Remainder Theorem we have Remainder R = f(k)

   In perfectly divisible case the remainder is zero(0).

Hence,f(k) = 0 for f(x) is exactly divisible by (x-k).

• Similarly, When f(k/a) = 0 ,then we can say that f(x) is exactly divisible by (ax-k).

Step to solve questions on remainder of algebraic function:-

Step-1: Equate the Divisor to zero(0) and find the value of x.

Step-2: Put that value of x in the polynomial given as dividend&the value will be remainder.

Q. Find the remainder when X³+5x²+7 is divided by (x-2) .

Sol:-  Equating x-2 to zero we get,

                        x-2 = 0 => x = 2

 Now,remainder will obtain by putting 2 in place of x in the dividend polynomial X³+5x²+7.

     R =  2³+5 x 2²+7

        = 8+20+7 = 35 (Ans)

Q. Find the remainder when X⁵¹+16 is divided by (x+1) .

Sol:- Equating x+1 to zero we get,

                        x+1 = 0 => x = -1

 Now,remainder will obtain by putting -1 in place of x in the dividend polynomial  X⁵¹+16.

     R = (-1)⁵¹+16

        = -1+16 = 15

Q. Find value of k for which,when  X²+4x+k is divided by (x-2) leaves remainder 2x .

SOL:-  Put x-2 = 0 => x = 2

             Remainder =  2²+4 x 2+k = 2x (Given)

                                => 4+8+k = 2 x 2 = 4

                               => k = -8

Q. Find value of  a for which  X³+ X² -5x+ a will be divisible by (x-2).

Sol:-  By factor theorem for x=2 the value of polynomial will be zero.

                 2³+ 2² -5 x 2+ a = 0

         => 8+4 -10+a = 0

        => a = -2

Hope,these concepts as well as techniques help you a lot to create your basics about finding remainder smartly which will in future help you to understand Remainder Theorems to solve more difficult problems.

 If you have any query then feel free to contact us & kindly give your valuable opinions in comment box for improvement of the present methods.And subscribe for getting new updates from this as it ain't over & continued.