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.

 

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