Remainder & Factor Theorems

2 Jan 1999

Polynomials
The Remainder Theorem
The Factor Theorem
Examples

olynomials

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

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

The constants a₀, a₁, ¼, a_{n -}₁, a_n are called the coefficients of xⁿ, xⁿ^{- 1}, ¼, x¹, x⁰ 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 (ax² + bx + c), then we may write

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:

(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:

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.