PreCalculus

3.4. Long Division

Given two polynomials, f(x) and d(x), if the degree of the divisor d(x) is less than or equal to the degree of the dividend f(x), then there is a unique polynomial quotient q(x) and a unique polynomial remainder r(x), such that:

f(x)          r(x)
---- = q(x) + ----
d(x)          d(x)

The degree of r(x) is less than the degree of d(x). If r(x) = 0, then d(x) is a factor of f(x). The process of performing such a division is called long division. Long division of polynomials can done in a manner similar to long division of real numbers. For example, to divide x² + 3x + 4 by x + 2 set up as follows:

      _____________
x + 2 )2x² + 3x + 4

Remove the first term of the dividend by multiplying the divisor by 2x and subtracting.

             2x    
x + 2 )2x² + 3x + 4
       2x² + 4x
            -1x + 4

Next, multiply the divisor by -1 and subtract from the remainder of the previous step:

             2x - 1   
x + 2 )2x² + 3x + 4
       2x² + 4x
            -1x + 4
            -1x - 2
                  6

Note that each column is in the same degree with the highest to the left. The result is:

2x² + 3x + 4              6
------------ = 2x - 1 + -----
    x + 2               x + 2

The binomial 2x - 1 is the polynomial quotient and 6 is the remainder. The degree of the remainder is less than the degree of the divisor. To check the result, multiply the quotient by the divisor and add the remainder:

(2x - 1)(x + 2) + 6 = 
2x² + 4x - x - 2 + 6 = 
2x² + 3x + 4

The remainder theorem can be derived from the polynomial quotient formula. Let,

d(x) = x - a

Then rewrite the formula as:

f(x) = (x - a) q(x) + r(x)

Set x = a, then

f(a) = (a - a)q(a) + r(a) = r

Remainder theorem: If a polynomial f(x) is divided by (x - a), the remainder is a constant given by f(a).

Example, determine the remainder of x³ + 2x² + 3x + 4 for the factor x + 2.

(-2)³ + 2(-2)² + 3(-2) + 4 =
-8 + 8 - 6 + 4 = -2

Now try another example, divide x³ - 8 by x - 2:

             x² + 2x + 4
x - 2 )x³ + 0x² + 0x - 8
       x³ - 2x²
            2x² + 0x - 8
            2x² - 4x
                  4x - 8
                  4x - 8
                       0

Terms with 0 a the coefficient were included in the dividend to make the long division process work out correctly. The result is:

x³ - 8
------ = x² + 2x + 4
 x - 2

Because the remainder is equal to zero, the polynomial has been factored into two polynomials of lower degree,

x³ - 8 = (x - 2)(x² + 2x + 4)

Applying the remainder theorem to this problem would give

f(2) = (2)³ - 8 = 8 - 8 = 0

The factor theorem is derived from the remainder theorem for a remainder of zero.

Factor theorem: For a polynomial f(x), (x - a) is a factor if and only if f(a) = 0.

Example, is (x + 2) a factor of f(x) = x² + 5x + 6?

f(-2) = (-2)² + 5(-2) + 6 = 4 - 10 + 6 = 0
Therefore, (x + 2) is a factor of f(x)