In the previous section you saw how to divide a polynomial by a factor. These ideas will be extended now to cover division of a polynomial by an expression of the form (x - a).
If (x - a) is not a factor, there will be a remainder. As with numbers, long divisions of polynomials often leave remainders.
For example (x + 3) is not a factor of x^3 + 6x^2 + 7x - 4, and so long division will yield a remainder.
| x^2 | + | 3x | – | 2 | |||
| x + 3 | x^3 | + | 6x^2 | + | 7x | - | 4 |
| x^3 | + | 3x^2 | |||||
| 3x^2 | + | 7x | |||||
| 3x^2 | + | 9x | |||||
| -2x | - | 4 | |||||
| -2x | - | 6 | |||||
| 2 | |||||||
One way of expressing this might be to write
\frac{x^3 + 6x^2 + 7x - 4}{x+3} = x^2 + 3x - 2,\ \text{rem} 2
but the normal method is to write either
x^3 + 6x^2 + 7x - 4 = (x^2 + 3x - 2) (x + 3) + 2
or
\frac{x^3 + 6x^2 + 7x - 4}{x + 3} = x^2 + 3x - \frac{2}{x+3}
The Factor Theorem can be extended to remainders. The Remainder Theorem is given below.
If P(x) is a polynomial of degree n and a is any number then
P(x) = (x - a) Q(x) + R
where Q(x) is a polynomial of degree (n - 1) and R = P(a).
Worked Examples
What is the remainder when x^3 - 7x + 10 is divided by (x - 1)?
Work it out by long division and check your answer using the Remainder Theorem.
| x^2 | + | x | – | 6 | |||
| x - 1 | x^3 | + | 0x^2 | - | 7x | + | 10 |
| x^3 | - | x^2 | |||||
| x^2 | - | 7x | |||||
| x^2 | - | x | |||||
| -6x | + | 10 | |||||
| -6x | + | 6 | |||||
| 4 | |||||||
Using the Remainder Theorem,
R = P(1) where P(x) = x^3 - 7x + 10
So
R = 1^3 - 7 × 1 + 10 = 4 (as expected)
Exercises
