In this section , we shall study a simple and an elegant method of finding the remainder.

Remainder Theorem

In the previous section , we have learnt the division of a polynomial by another non – zero polynomial.

In this section , we shall study a simple and an elegant method of finding the remainder.

In the case of divisibility of a polynomial by a linear polynomial we use a well known theorem called Remainder Theorem.

If a polynomial p(x) of degree greater than or equal to one is divided by a linear polynomial (x–a) then the remainder is p(a), where a is any real number.

Significance of Remainder theorem : It enables us to find the remainder without actually following the cumbersome process of long division.

It leads to another well known theorem called ‘Factor theorem’.

If the polynomials *f(x)* = *ax*3 + 4*x* 2 + 3*x* –4 and *g(x)* = *x*3– 4*x* + *a* leave the same remainder when divided by *x*–3, find the value of *a*. Also find the remainder.

*Solution*

Let *f(x)* = *ax*3 + 4*x* 2 + 3*x* –4 and *g(x)* = *x*3– 4*x* + *a*, When f(x) is divided by (x–3), the remainder is *f*(3).

Now *f(x)* = *a*(3)3 + 4(3) 2 + 3(3) –4

= 27a + 36 + 9 – 4

*f*(3) = 27a + 41 (1)

When g(x) is divided by (x–3), the remainder is g(3).

Now g(3) = 33 – 4(3) + a

= 27 – 12+ a

= 15 + a (2)

Since the remainders are same, (1) = (2)

Given that, *f*(3) = *g*(3)

That is 27a + 41 = 15 + a

27a – a = 15 – 41

26a = –26

a = - 26/26 = –1

Substituting a = –1,in f(3), we get

f(3) = 27( ) - + 1 14

= – 27 + 41

f(3) = 14

so The remainder is 14.

Without actual division , prove that *f(x)* = 2 *x*4 - 6* x*3 + 3* x*2 + 3*x* - 2 is exactly divisible by *x*2 –3*x* + 2

*Solution :*

*Let f(x)* = 2 *x*4 - 6* x*3 + 3* x*2 + 3*x* - 2

*g(x) = x*2 –3*x* + 2

= *x*2-2*x*-*x*+2

=*x*(*x*-2)-1(*x*-2)

=(*x*-2)(*x*-1)

we show that *f(x)* is exactly divisible by (*x*–1) and (*x*–2) using remainder theorem

*f(*1)= *2 *(1)*4 - 6 *(1)*3 + 3 *(1)*2 + 3*(1)* - 2*

*f(*1)=2-6+3+3-2=0

*f(*2)=* 2 *(2)*4 - 6 *(2)*3 + 3 *(2)*2 + 3*(2)* - 2*

*f(*2)=32-48+12+6=0

*f(x*) is exactly divisible by (*x* – 1) (*x* – 2)

i.e., *f(x)* is exactly divisible by *x*2 –3*x* + 2

If
*p*(*x*) is divided by (*x* -*a*) with the remainder *p*(*a* ) =
0 , then (*x* -*a*) is a factor of *p*(*x*). Remainder Theorem leads
to Factor Theorem.

** **

__1. Factor Theorem__

If *p*(*x*)
is a polynomial of degree *n* ≥ 1 and ‘*a*’ is any real number
then

(i) *p*(*a
*)* *=* *0* *implies* *(*x *-*a*)* *is a factor of* p*(*x*).

(ii) (*x*
-*a*) is a factor of *p*(*x*) implies
*p*(*a* ) = 0 .

**Proof**

If *p*(*x*)
is the dividend and (*x* -*a*) is a divisor, then by division algorithm we write, *p*(*x*)
=
(*x* −*a*)*q* (*x*) + *p*(*a*) where *q* (*x*)
is the quotient and *p*(*a*) is the remainder.

(i)
If *p*(*a*) = 0, we get *p*(*x*) =
(*x* −*a*)*q* (*x*) which shows that (*x* -*a*) is a factor of *p*(*x*).

(ii)
Since (*x* -*a*) is a factor of *p*(*x*) , *p*(*x*) =
(*x* – *a*)*g *(*x*)* *for some polynomial*
g *(*x*)* *.

In
this case

*p*(*a *)* *=*
*(*a *−*a*)*g *(*a *)

= 0
×*g* (*a* )

=
0

Hence,
*p*(*a*) = 0, when (*x* -*a*) is a factor of *p*(*x*).

**Thinking Corner**

For any two integers *a*(*a* ≠ 0) and *b, a* divides *b* if* b *=*
ax*, for some integer* x.*

**Note**

• (*x* -a) is a factor of *p*(*x*) , if *p(a*) = 0 ( *x–*a
= 0,* x *= a)

• (*x +* a) is a factor of
*p*(*x*)
, if *p*(–*a*) = 0 ( *x+a* = 0,* x *= –*a*)

• (*ax*+*b*) is a factor of *p*(*x*) , if *p*(–*b/a*)
= 0 ( a*x + b *= 0, a*x =* −b,* x *= − *b/a*)

• (a*x–*b) is a factor of
*p(x)* , if *p*(*b/a*) = 0 { (a*x*
–* b*)* *= 0, a*x =* b,* x *= *b/a*
)

• (*x–a*) (*x–b*) is a factor of *p*(*x*) , if *p*(*a*)=0
and *p*(*b*) = 0 ( * x *− a = 0 or* x *−* b *= 0 , *x *= a or * x *=* b *)

**Example 3.13**

Show
that (*x* + 2) is a factor of *x* ^{3}
−
4*x*^{2} −2*x* +
20

*Solution*

Let*
p*(*x*)* *=* x *^{3}* *−* *4*x*^{2}* *−*
*2*x *+*
*20

To find the zero of *x*+2;

put *x* + 2 = 0

we get *x* = –2

By
factor theorem, (*x +* 2) is factor of *p(x*), if *p*(− 2) = 0

* p*(−
2) = (− 2)^{3} − 4(− 2)^{2} −2(− 2) + 20

= −8−4(4)+4+20

*p*(−* *2)* *=* *0

Therefore,
(*x* + 2) is a factor of *x*^{3} −
4*x*^{2} −2*x* +
20

**Example 3.14**

Is
(3*x* -2) a factor of 3*x*^{3} +
*x*^{2} −20*x* +12
?

*Solution*

Let*
p*(*x*)* *=* *3*x*^{3}* *+* x*^{2}* *−20*x *+12

Therefore,(3*x*
-2)
is a factor of

3*x*
^{3} + *x*^{2} −20*x* +12

**Progress Check**

1. (*x*+3) is a factor of *p(x)*,
if *p(*__) = 0

2. (3*–x*) is a factor of *p(x)*, if
*p(*__) = 0

3. (*y–*3) is a factor of *p(*y),
if *p(*__) = 0

4. (–*x–*b) is a factor of *p(x)*,
if *p(*__) = 0

5. (*–x*+b) is a factor of *p(x)*,
if *p(*__) = 0

**Example 3.15**

Find the
value of *m*, if (*x* -2) is a factor of the polynomial 2*x*^{3}
−
6*x*^{2} +*mx* + 4 .

*Solution*

Let
*p*(*x*) = 2*x*^{3} −6*x*^{2}
+
*mx* + 4

To find the zero of *x*–2;

put* x *– 2 = 0

we get *x =* 2

By
factor theorem, (*x* -2) is a factor of *p*(*x*)
, if *p*(2) = 0

*p*(2) = 0

2(2)^{3}
−
6(2)^{2} +*m*(2) + 4 = 0

2(8)
−
6(4) +
2*m* + 4 = 0

− 4 + 2*m* =
0

*m
*= 2

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