The Remainder Theorem

 

If you are doing a long division problem such as 683 ¸22, you would write it like this:

and so you would write that 683 ¸22 = 31 with remainder 1 or you would write 683 = (22)(31) + 1.  As a matter of fact the latter way is the way you would check your answer.  In this problem 683 is called the dividend, 22 is called the divisor, 31 is called the quotient, and 1 is called the remainder.

 

Well, the same thing happens when you have long division of polynomials.  Let p(x) be the dividend, d(x) be the divisor, q(x) be the quotient and r(x) be the remainder.  So

p(x) = d(x)q(x) + r(x).

 

 Now suppose that the thing you are dividing by (d(x)) is a simple expression like x – c.  Then you have p(x) = (x – c)q(x) + r.  (The remainder is a constant, since it always has degree less than the divisor and the divisor has degree one.)  Now what happens when you calculate p(c)?  p(c) =(c – c)q(c) + r = 0(q(c)) + r = 0 + r = r.  So p(c) = r.  Now, before your eyes glaze over in confusion think about what this is saying:  “You get the same result when you evaluate the polynomial at x = c, as the remainder when you divide by the expression x - c.

 

Still confused?  Well maybe and an example will help out.

Let the polynomial p(x) = 5x³ - 2x – 1.  If you divide p(x) by x – 3, you get (synthetically):

i.e. (5x3 - 2x – 1) ¸ (x – 3) = (5x2 +15x + 43) with remainder 128. Altrnatively: 5x3 - 2x – 1 = (x – 3) (5x2 +15x + 43) + 128 So p(3) = 128.  Check it out.

 

This theorem is used two ways:  You can calculate the remainder when dividing by x –c, by evaluating p(c) or you can calculate p(c ) by dividing by x – c.  Always do whichever calculation is simplest.  If you are getting bogged down in a tedious calculation, you are probably not using the theorem correctly.