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’.

Example: 3.15

Example 3.16

If the polynomials f(x) = ax³ + 4x ² + 3x –4 and g(x) = x³– 4x + a leave the same remainder when divided by x–3, find the value of a. Also find the remainder.

Solution

Let f(x) = ax³ + 4x ² + 3x –4 and g(x) = x³– 4x + 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.

Example 3.17

Without actual division , prove that f(x) = 2 x⁴ - 6 x³ + 3 x² + 3x - 2 is exactly divisible by x² –3x + 2

Solution :

Let f(x) = 2 x⁴ - 6 x³ + 3 x² + 3x - 2

g(x) = x² –3x + 2

= x²-2x-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 x² –3x + 2