# Remainder and Factor Theorems

## Remainder and Factor Theorems

Remainder Theorem: The remainder theorem states that if a polynomial f(x) is divided by a linear function x – k, then the remainder is f(k).

Proof: In any division

Divided = divisor x Quotient + Remainder

Let Q(x) be the quotient and R be the remainder

Then f (x) = (x – k) Q (x) + R

f (x) = (k – k) Q (x) + R = 0 + R = R

Note: If an n – degree polynomial is divided by an m degree polynomial. Then the maximum degree of the remainder polynomial is m – 1

Example: Find the remainder, when x³ + 4x² – 7x + 6 is divided by x – 1

Solution: Let f(x) = x³ + 4x² – 7x + 6. The remainder when f(x) is divided by x – 1 is

x – 1 = 0

x = 1

f(1) = (1)³ + 4(1)² –  7(1) + 6

= 1 + 4 – 7 + 6

= 4

Factor Theorem: Factor theorem is a special case of remainder theorem

Let f (x) = (x – k) Q (x) + R

f (x) = (x – k) Q (x) + f (k)

When f(k) = 0, f(x) = (x – k) Q(x). therefore, f(x) is exactly divisible by x – k

Example: x² + x – 6 is a factor of 2x⁴ + x³ – ax² + bx + a + b – 1, find the values of a and b.

Solution: Given that x² + x – 6 = 0
x² + 3x – 2x – 6 = 0

x (x + 3) – 2(x + 3) = 0

(x + 3) (x – 2) = 0

x + 3 = 0

x = -3

and x – 2 = 0

x = 2

Let f(x) = 2x⁴ + x³ – ax² + bx + a + b – 1 = 0

Now

f(-3) = 2(-3)⁴ + (-3)³ – a(-3)² + b(-3) + a + b – 1 = 0

134 – 8a – 2b = 0

4a + b = 67 … (1)

Now

f(2) = 2(2)⁴ + (2)³ – a(2)² + b(2) + a + b – 1 = 0

39 – 3a + 3b = 0

a – b = 13 … (2)

from equation (1) and (2)

a = 16 and b = 3.