September 10, 2024

# Remainder Theorem

Remainder theorem

# Zeros of polynomials:

The value of a polynomial      at

is  obtained by putting    in
and is denoted by  .

e.g.  Find the value of    at

.

Sol. On  putting    in given polynomial , we get

This implies  .

Thus value of

at    is 12.

We say that a zero of a polynomial

is a number  such that
.  It means is  zero of  a polynomial
if the value of that polynomial at  is  zero.

The zeros of polynomials is obtained by  equating  the given polynomial

.  We say $p\left&space;(&space;x&space;\right&space;)=0$ is a polynomial equation and $c$
is root of the polynomial equation $p\left&space;(&space;x&space;\right&space;)=&space;0$.

Let  $p\left&space;(&space;x&space;\right&space;)=a$

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

Now replace $x$

with any number  we still get $p\left&space;(&space;x&space;\right&space;)=a$. 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

be a linear polynomial,

then      means

.

So

is the only zero of  .

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

are -2 and 2.

(iii) 0 may be a zero of polynomial.

e.g.  Take

(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  $0\leq&space;remainder<&space;divisor$

.  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        $(x-2)(3x^{3}+2x^{2}+4x+5)+8=&space;3x^{4}+2x^{3}+4x^{2}+5x-6x^{3}-4x^{2}-8x-10+8$

$\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;=&space;3x^{4}-4x^{3}-3x-2$

$\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;=p(x)$

Hence  $p(x)=g(x)\times&space;q(x)+r(x)$  where  $deg(r(x))=0<&space;1=deg(g(x))$

.

In general , If $p(x)$ and $g(x)$

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

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 $\large&space;a$

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

$\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;\,&space;p(x)=(x-a)q(x)+r(x)$

since degree of $r(x)<$ degree of $q(x)$

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

$\Rightarrow&space;r(x)=c$

(a constant polynomial)

$\Rightarrow&space;p(x)=(x-a)q(x)+c$

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.