2 Jan 1999

### olynomials

A polynomial of degree n in x is an expression of the form

P(x) = a0xn + a1xn - 1 + a2xn - 2 + ¼ + an - 1x + an
where a0, a1, ¼, an - 1an are constants with a0 ¹ 0, and n ³ 0 is an integer.

The constants a0, a1, ¼, an - 1an are called the coefficients of xn, xn - 1, ¼, x1, x0 respectively.

1. If a polynomial f(x) is divided by a linear divisor (ax + b), then we may write
2. f(x) = (ax + b)Q(x) + R,

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

3. If a polynomial f(x) is divided by a quadratic divisor (ax2 + bx + c), then we may write
4. f(x) = (ax2 + 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 2x3 + ax2 + 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) = 2x3 + ax2 + 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) = 2x3 + x2 - 5x + 2.

Since (x + 2) is a factor of f(x), by long division or inspection,
f(x) = (x + 2)(2x2 - 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 x3 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 x3 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.

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