Remainder Theorem

Zeros of polynomials:

The value of a polynomial    p(x)  at  x=a  is  obtained by putting  x=a  in p(x) and is denoted by  .

e.g.  Find the value of  p(x)=x^{2}-5x+6  at  x=-1.

Sol. On  putting  x=-1  in given polynomial , we get

p(-1)=(-1)^{2}-5(-1)+6=1+5+6

This implies  p(-1)=12.

Thus value of p(x) at  x=-1  is 12.

 We say that a zero of a polynomial  p(x) is a number \large c such that p(c)=0.  It means \large c is  zero of  a polynomial  p(x)  if the value of that polynomial at \large c is  zero.

The zeros of polynomials is obtained by  equating  the given polynomial  p(x)=0.  We say p\left ( x \right )=0 is a polynomial equation and c is root of the polynomial equation p\left ( x \right )= 0.

Let  p\left ( x \right )=a be a constant polynomial, then p\left ( x \right )=ax^{0}.

Now replace x  with any number  we still get p\left ( x \right )=a. This implies constant polynomials has no zeros.  In case of zero polynomial , every real number is a zero of the zero polynomial.

Important observations:

(i) Every linear polynomial has one and only one zero.

Let     \large p(x)=ax+b, \, a\neq 0   be a linear polynomial,

then    p(x)=0  means   ax+b=0.

                                                \Rightarrow x=-\frac{b}{a}

So   x=-\frac{b}{a}    is the only zero of  p(x).

i.e.  a linear  linear polynomial has one and only one zero.

(ii) A zero of polynomial need not be 0.

e. g.  The zeros of     p(x)=x^{2}-4    are -2 and 2.

(iii) 0 may be a zero of polynomial.

e.g.  Take    p(x)=x(x-2)

(iv) A polynomial can have  more than one zero. 

 

Division algorithm in polynomials:

When we divide two numbers, we always get

Dividend =(divisor x quotient)+remainder,   where  0\leq remainder< divisor.  When remainder becomes zero, we say divisor and quotient both are factors of dividend.

Now, let two polynomials p(x)=3x^{4}-4x^{3}-3x-2  and g(x)=x-2 . Divide p(x) by g(x) 

 

{ 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, until the degree of the new dividend is less than the degree of the divisor.} 

Now        (x-2)(3x^{3}+2x^{2}+4x+5)+8= 3x^{4}+2x^{3}+4x^{2}+5x-6x^{3}-4x^{2}-8x-10+8

                                                              \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 3x^{4}-4x^{3}-3x-2

                                                              \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =p(x)

 Hence  p(x)=g(x)\times q(x)+r(x)  where  deg(r(x))=0< 1=deg(g(x)).

In general , If p(x) and g(x) are two polynomials  such that deg(p(x))\geq deg(g(x))  and g(x)\neq 0, then we can find a polynomial q(x) as quotient  and r(x) as remainder, where r(x)=0  or deg(r(x))< deg(g(x)).

In the above example the divisior is a linear polynomial . In such a situation there is a way to find the  remainder called Remainder Theorem.

Remainder theorem:  Let p(x) be any polynomial of degree greater than or equal to one  and let \large a be any real number. If  p(x) is divided by the  linear polynomial  x-a, then remainder is  p(a).

Proof. Let p(x) be any polynomial of degree greater than or equal to 1. Suppose p(x) is divided by (x-a) then by using division algorithm theorem , p(x) can be written as

\, \, \, \, \, \, \, \, \, \, \, \, \, \, p(x)=(x-a)q(x)+r(x)

since degree of r(x)< degree of q(x),  this implies degree of r(x)=0  ( \because degree of q(x)=1)

                                              \Rightarrow r(x)=c   (a constant polynomial)

                                            \Rightarrow p(x)=(x-a)q(x)+c

In particular if x=a  then  p(a)=c,  which proves the theorem. 

e.g. Find the remainder when x^{5}+x^{4}-2x^{3}+x^{2}+1 is divided by (x-1).

Sol. Here p(x)=x^{5}+x^{4}-2x^{3}+x^{2}+1 and zeros of x-1 is 1. 

so p(1)=(1)^{5}+(1)^{4}-2(1)^{3}+(1)^{2}+1

              =1+1-2+1+1

              =2

Hence by the remainder theorem , the remainder is 2.

 

 

 

                                                                

 

 

 

 

 

 

 

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