Home | Remainder Theorem

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

**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

*g(x) = x*

= *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)

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

*f(*2)=* 2 *(2)* ^{4} - 6 *(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

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