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) = 3x2 -2x + 4 by the polynomial P(x) = x + 2
Since P1 is being divided by P2 it is called the dividend while P2 is called the divisor. The result of division of P1 by P2 is called the quotient and whatever is left after division is called the remainder.
The procedure of division of P1 by P2 can best be demonstrated step by step as follows:
Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.
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) = x3 + 2x3 – 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) = -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.
If the polynomial f(x) is divided by x – a, the remainder is f(a).
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
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).
Given that f(x) = 3x3 – 4x2 + 2x + 3, find the remainder when f(x) is divided by x – 1.
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
Find the remainder when
f(x) = (x + 3)(x – 2)(x + 2) is divided by x + 1.How Can We Make ClassNotesNG Better - CLICK to Tell Us💃
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