Polynomials The Remainder Theorem The Factor Theorem Examples

olynomials

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

The constants a₀, a₁, ¼, a_{n - 1},  a_n are called the coefficients of xⁿ, x^{n - 1}, ¼, x¹, x⁰ respectively.

1. 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.

2. 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

actor Theorem

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

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.