**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*

*Workout Examples*

*Example 1: Find the remainder when x ^{3}
– 4x^{2} + 8x – 5 is divided by x – 2.*

*Solution:** Here,*

* f(x) = x ^{3} –
4x^{2} + 8x – 5 *

* Comparing x – 2 with x
– a, a = 2*

* **∴** by remainder theorem,*

* Remainder
R = f(2)*

* = 2 ^{3} - 4×2^{2} + 8 ×
2 – 5 *

* = 8 – 16 + 16 – 5*

* = 3*

*Example 2: Find the remainder when 4x ^{3}
+ 2x^{2} – 4x + 3 is divided by 2x + 3.*

*Solution:** Here,*

* f(x) = 4x ^{3} +
2x^{2} – 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 x ^{3}
– 3x^{2} + mx + m is divided by x + 2, a remainder -2m is obtained,
find the value of m.*

*Solution:** Here,*

* f(x) = x ^{3} – 3x^{2}
+ 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.*

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