Back to: Further Mathematics SS2

**Division of Polynomials**

A polynomial of degree *n *can be divided by another polynomial of degree *m* if *n ≥ m.*

Divide the polynomial P(x) = 3x^{2} -2x + 4 by the polynomial P(x) = x + 2

**Solution**

Since P_{1} is being divided by P_{2} it is called the dividend while P_{2} is called the divisor. The result of division of P_{1} by P_{2} is called the quotient and whatever is left after division is called the remainder.

The procedure of division of P_{1} by P_{2} can best be demonstrated step by step as follows:

**Step 1**

Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.

Now, find:

a. f(1)

b. f(2)

c. f(3)

d. f(1) = 2 – 3 + 4 – 1 = 2

e. f(2) = 16 – 12 + 8 – 1 = 11

f. f(3) = 54 – 27 + 12 – 1 = 38

Compare the answers in each of the following pairs

i. (a) and (d)

ii. (b) and (e)

iii. (c) and (f)

What do you notice?

This is a remarkable result and it is not a mere coincidence.

Let us try another one.

Given that H(x) = x^{3} + 2x^{3} – 4x + 3, find the remainder when H(x) is divided by

(a) x + 1

(b) x + 2

(c) x + 3

When H(x) is divided by x + 3, the remainder is 6. Now, find:

(d) H(-1)

(e) H(-2)

(f) H(-3)

(d) H(-1) = -1 + 2 + 4 + 3 = 8

(e) H(-2) = -8 + 8 + 8 + 3 = 11

(f) H(-3) = -27 + 18 + 12 + 3 = 6

Compare again the answers in each of the following pairs:

(i) (a) and (d)

(ii) (b) and (e)

(iii) (c) and (f)

What again do you notice?

(a) When H(x) is divided by x + 1 the remainder is H(-1)

(b) When H(x) is divided by x + 2 the remainder is H(-2)

(c) When H(x) is divided by x + 3 the remainder is H(-3)

So we can safely conclude that if H(x) is divided by x – a the remainder is H(a) and this forms the basis of what is called the remainder theorem.

**Theorem**

If the polynomial f(x) is divided by x – a, the remainder is f(a).

*Proof*

The polynomial function f(x) can be written as:

f(x) = (x – a) Q(x) + R …(1)

where x – a is the divisor, Q(x) the quotient and R the remainder.

Put x = a in (1)

f(a) = (a – a) Q(a) + R

f(a) = R

The theorem whose proof we have just established, is called the **RemainderTheorem**. In general, if f(x) is divided by ax + b then the remainder is

f

A special case of the remainder theorem is when f(x) leaves no remainder when it is divided by x – a. We therefore say that x – a is a factor of f(x). The modified theorem is called, **Factor Theorem** and it states:

If f(a) = 0 then x – a is a factor of f(x).

**Example**

Given that f(x) = 3x^{3} – 4x^{2} + 2x + 3, find the remainder when f(x) is divided by x – 1.

**Solution**

Let R be the remainder. Then using the remainder theorem, R = f(1)

f(1) = 3(1)^{3} – 4(1)^{2} + 2(1) +3

= 3 – 4 + 2 + 3

= 4

Find the remainder when

f(x) = (x + 3)(x – 2)(x + 2) is divided by x + 1.

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