where a_{0}, a_{1}, ¼,
a_{n - }_{1}, a_{n}
are constants with a_{0} ¹
0, and n ³ 0 is an integer.

The constants a_{0}, a_{1}, ¼,
a_{n - }_{1}, a_{n}
are called the coefficients of x^{n}, x^{n}^{
- 1}, ¼,
x^{1}, x^{0} respectively.

If a polynomial f(x) is divided by a linear divisor (ax
+ b), then we may write

f(x) = (ax
+ b)Q(x) + R,

where Q(x) is the quotient and R is the
remainder.

If a polynomial f(x) is divided by a quadratic divisor (ax^{2} + bx + c), then we may write

f(x) = (ax^{2} + bx + c)Q(x) + (Rx + S),

where Q(x) is the quotient and (Rx + S) is the remainder.

emainder Theorem

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

In general, if f(x) is divided by (ax - b),
then the remainder is f(b/a).

actor Theorem

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

Conversely, if f(a) = 0, then (x
- a) is a factor of f(x).

xamples

Example 1: The expression 2x^{3}
+ ax^{2} + bx + 2 is exactly divisible by (x
+ 2) and leaves a remainder of 12 on division by (x -
2). Calculate the values of a and b and factorise the
expression completely.

Solution:

Let f(x) = 2x^{3} + ax^{2}
+ bx + 2.

(x + 2) is a factor of f(x), so f(-2)
= 0.
f(-2) = 2(-2)^{3}
+ a(-2)^{2} + b(-2)
+ 2 = -16 + 4a -
2b + 2 = 0,
ie, 4a - 2b
= 14.

Division by (x - 2) leaves a remainder
of 12, so f(2) = 12.
f(2) = 2(2)^{3} + a(2)^{2} + b(2)
+ 2 = 16 + 4a + 2b + 2 = 12,
ie, 4a + 2b = -6.

Solving gives a = 1 and b = -5.
\ f(x) = 2x^{3}
+ x^{2} - 5x + 2.

Since (x + 2) is a factor of f(x), by long division
or inspection,
f(x) = (x + 2)(2x^{2} -
3x + 1)
f(x) = (x + 2)(2x -
1)(x - 1)

Example 2: When the polynomial P(x)
is divided by (x - 1) the remainder is
7, and when divided by (x - 3) the remainder
is 13. Find, by writing

P(x) = (x -
1)(x - 3)Q(x) + ax
+ b,

the remainder when P(x) is divided
by (x - 1)(x -
3). If P(x) is a cubic in which the coefficient of
x^{3} is unity and P(2) = 6, determine Q(x).

Solution:

P(1) = 7 Þ a + b
= 7.
P(3) = 13 Þ 3a + b
= 13.

Solving gives a = 3, b = 4.
\P(x) = (x - 1)(x
- 3)Q(x) + 3x + 4.
Thus the remainder on division by (x -
1)(x - 3) is 3x + 4.

If P(x) is a cubic with coefficient of x^{3}
unity, then Q(x) = (x + c),
ie, P(x) = (x - 1)(x
- 3)(x + c) + 3x + 4.
P(2) = 6 Þ c = 2.
\ Q(x) = x + 2.