Example 1:

Show that 2 is a root of p(x) = x3 - 3x2 + x + 2 and factor as far as possible.

Solution:
First we calculate p(2).

p(2) = (2)3 - 3(2)2 + (2) + 2 = 8 - 12 + 2 + 2 = 0

So p(x) has (x - 2) as one of the factors. Now we use long division to divide (x-2) into p(x).

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Thus we get p(x) = (x - 2)(x2 - x - 1).

Now by factoring (x2 - x - 1) like we did before, we get

x2 - x - 1 = [x - (1 + SQRT(5))/2]*[x - (1 - SQRT(5))/2]

Hence we get that p(x) = (x - 2)[x - (1 + SQRT(5))/2]*[x - (1 - SQRT(5))/2]
Example 2

What if the term that you are dividing is not a factor of the polynomial? You can still do the long division but you get a remainder. Here is an example:

Divide x-2 into x3 - x2 + 3x - 7.

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So we conclude that x3 - x2 + 3x - 7 = (x - 2)(x2 + x + 5) + 3

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