1. Remainder Theorem

2. When a polynomial f(x) is divided by x – a, we can find the remainder as follows:
Let Q(x) and R be the quotient and the remainder respectively.
By division algorithm, we have
Since the degree of R is less than that of x – a, the degree of R is 0. So, R is a constant. 
f(x) = (x – a)  Q(x) + R
When x = a, the value of the polynomial is
f(a) = (a – a)  Q(a) + R
= 0  Q(a) + R
= R
∴ Remainder = f(a)

3. In conclusion, we have the following theorem:
Theorem 5.1 Remainder theorem
When a polynomial f(x) is divided by x – a, the remainder is equal to f(a).
When f(x) = x2 + 2x – 1 is divided by x – 1,
remainder = f(1)
= (1)2 + 2(1) – 1
= 2

4. In conclusion, we have the following theorem:
Theorem 5.1 Remainder theorem
When a polynomial f(x) is divided by x – a, the remainder is equal to f(a).
When f(x) = x2 + 2x – 1 is divided by x + 2,
x + 2 = x – (–2)
remainder = f(–2)
= (–2)2 + 2(–2) – 1
= –1

5. Follow-up question Find the remainder when the polynomial
x3 – 2x2 – 3x + 5 is divided by x + 1.
Let f(x) = x3 – 2x2 – 3x + 5.
By the remainder theorem,
x + 1 = x – (–1)
remainder = f(–1)
= (–1)3 – 2(–1)2 – 3(–1) + 5
= –1 –
= 5

6. What is the remainder when a polynomial f(x) is divided by mx – n?
The remainder theorem can be extended as follows:
Theorem 5.2
When a polynomial f(x) is divided by mx – n, the
remainder is equal to | ø ö ç è æ m n f

7. Find the remainder when 9x2 + 6x – 7 is divided by 3x + 1.
Let f(x) = 9x2 + 6x – 7.
By the remainder theorem,
remainder = | ø ö ç è æ – 3 1 f
3x + 1 = 3x – (–1) 7 3 1 6 9 2 – | ø ö ç è æ + =
= 1 – 2 – 7
= –8

8. Follow-up question
When 4x3 + kx2 – x + 2 is divided by 4x – 1, the remainder is 2. Find the value of k.
Let f(x) = 4x3 + kx2 – x + 2.
By the remainder theorem, 3 k = 16