# Remainder Theorem

## Remainder Theorem

The remainder theorem states that - when a polynomial f(x) is divided by a binomial x – a, then the remainder is f(a).

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Proof: If we divide a polynomial f(x) by a binomial x – a, then we get Q(x) as quotient and R as remainder. Then,

f(x) = (x – a).Q(x) + R

Putting x = a, then

f(a) = (a – a).Q(x) + R

or,        f(a) = 0 + R

or,        R = f(a)

Hence, the remainder R = f(a). Proved.

Corollary 1: If a polynomial f(x) is divided by (x + a), then the remainder R = f(-a).

Proof: If we divide f(x) by (x + a), then we get Q(x) as quotient and R as remainder. Then,

f(x) = (x + a).Q(x) + R

Putting x = -a, then

f(-a) = (-a + a).Q(x) + R

or,        f(-a) = 0 + R

or,        R = f(-a)

Hence, the remainder R = f(-a). Proved

Corollary 2: If a polynomial f(x) is divided by (ax + b), then the remainder R = f(-b/a).

Proof: When we divide f(x) by ax + b, then we get Q(x) as quotient and R as remainder. Then,

f(x) = (ax + b).Q(x) + R = a(x + b/a).Q(x) + R

Putting x = -b/a, then

f(-b/a) = a(-b/a + b/a).Q(x) + R

or,        f(-b/a) = 0 + R

or,        R = f(-b/a)

Hence, the remainder R = f(-b/a). Proved.

Corollary 3: If a polynomial f(x) is divided by (ax – b), then the remainder R = f(b/a).

Proof: When we divide f(x) by ax – b, then we get Q(x) as quotient and R as remainder. Then,

f(x) = (ax – b).Q(x) + R = a(x – b/a).Q(x) + R

Putting x = b/a, then

f(b/a) = a(b/a – b/a).Q(x) + R

or,        f(b/a) = 0 + R

or,        R = f(b/a)

Hence, the remainder R = f(b/a). Proved.

### Workout Examples

Example 1: Find the remainder when x3 – 4x2 + 8x – 5 is divided by x – 2.

Solution: Here,

f(x) = x3 – 4x2 + 8x – 5

Comparing x – 2 with x – a, a = 2

by remainder theorem,

Remainder R = f(2)

= 23 - 4×22 + 8 × 2 – 5

= 8 – 16 + 16 – 5

= 3

Example 2: Find the remainder when 4x3 + 2x2 – 4x + 3 is divided by 2x + 3.

Solution: Here,

f(x) = 4x3 + 2x2 – 4x + 3

Divisor 2x + 3 can be written as 2(x + 3/2) and comparing with x – a, a = -3/2

by remainder theorem,

Remainder R = f(-3/2)

= 4×(-3/2)3 + 2×(-3/2)2 4 × (-3/2) + 3

= 4×-27/8 + 9/2 + 6 + 3

= -27/2 + 9/2 + 9

= 0

Example 3: When a polynomial x3 – 3x2 + mx + m is divided by x + 2, a remainder -2m is obtained, find the value of m.

Solution: Here,

f(x) = x3 – 3x2 + mx + m

Comparing x + 2 with x – a, a = -2

by remainder theorem,

f(-2) = Remainder

i.e        (–2)3 3×(–2)2 + m × (–2) + m = 2m

or,        –20 – m  = –2m

or,        –m + 2m = 20

or,        m = 20

The value of m = 20.

You can comment your questions or problems regarding the remainder theorem here.